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January 27, 2005 11:45 L24-CH08 Sheet number 1 Page number 337 black 337 CHAPTER 8 Principles of Integral Valuation EXERCISE SET 8.1 1. u =4 2x, du = 2dx, 1 2 u 3 du = 1 8 u 4 + C = 1 8 (4 2x) 4 + C 2. u =4+2x, du =2dx, 3 2 u du = u 3/2 + C = (4 + 2x) 3/2 + C 3. u = x 2 , du =2xdx, 1 2 sec 2 u du = 1 2 tan u + C = 1 2 tan(x 2 )+ C 4. u = x 2 , du =2xdx, 2 tan u du = 2 ln | cos u | + C = 2 ln | cos(x 2 )| + C 5. u = 2 + cos 3x, du = 3 sin 3xdx, 1 3 du u = 1 3 ln |u| + C = 1 3 ln(2 + cos 3x)+ C 6. u = 2 3 x, du = 2 3 dx, 1 6 du 1+ u 2 = 1 6 tan 1 u + C = 1 6 tan 1 2 3 x + C 7. u = e x , du = e x dx, sinh u du = cosh u + C = cosh e x + C 8. u = ln x, du = 1 x dx, sec u tan u du = sec u + C = sec(ln x)+ C 9. u = tan x, du = sec 2 xdx, e u du = e u + C = e tan x + C 10. u = x 2 , du =2xdx, 1 2 du 1 u 2 = 1 2 sin 1 u + C = 1 2 sin 1 (x 2 )+ C 11. u = cos 5x, du = 5 sin 5xdx, 1 5 u 5 du = 1 30 u 6 + C = 1 30 cos 6 5x + C 12. u = sin x, du = cos x dx, du u u 2 +1 = ln 1+ 1+ u 2 u + C = ln 1+ 1 + sin 2 x sin x + C 13. u = e x , du = e x dx, du 4+ u 2 = ln u + u 2 +4 + C = ln e x + e 2x +4 + C 14. u = tan 1 x, du = 1 1+ x 2 dx, e u du = e u + C = e tan 1 x + C 15. u = x 1, du = 1 2 x 1 dx, 2 e u du =2e u + C =2e x1 + C 16. u = x 2 +2x, du = (2x + 2)dx, 1 2 cot u du = 1 2 ln | sin u| + C = 1 2 ln sin |x 2 +2x| + C 17. u = x, du = 1 2 x dx, 2 cosh u du = 2 sinh u + C = 2 sinh x + C

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### Transcript of Calculus Early Transcendentals 8th Edition Solution Manual Chapter_08

January 27, 2005 11:45 L24-CH08 Sheet number 1 Page number 337 black

337

CHAPTER 8

Principles of Integral Valuation

EXERCISE SET 8.1

1. u = 4− 2x, du = −2dx,−12

∫u3 du = −1

8u4 + C = −1

8(4− 2x)4 + C

2. u = 4 + 2x, du = 2dx,32

∫ √u du = u3/2 + C = (4 + 2x)3/2 + C

3. u = x2, du = 2xdx,12

∫sec2 u du =

12tanu+ C =

12tan(x2) + C

4. u = x2, du = 2xdx, 2∫tanu du = −2 ln | cosu |+ C = −2 ln | cos(x2)|+ C

5. u = 2 + cos 3x, du = −3 sin 3xdx, − 13

∫du

u= −1

3ln |u|+ C = −1

3ln(2 + cos 3x) + C

6. u =23x, du =

23dx,

16

∫du

1 + u2 =16tan−1 u+ C =

16tan−1 2

3x+ C

7. u = ex, du = exdx,∫sinhu du = coshu+ C = cosh ex + C

8. u = lnx, du =1xdx,

∫secu tanu du = secu+ C = sec(lnx) + C

9. u = tanx, du = sec2 xdx,∫eu du = eu + C = etan x + C

10. u = x2, du = 2xdx,12

∫du√1− u2

=12sin−1 u+ C =

12sin−1(x2) + C

11. u = cos 5x, du = −5 sin 5xdx, − 15

∫u5 du = − 1

30u6 + C = − 1

30cos6 5x+ C

12. u = sinx, du = cosx dx,∫

du

u√u2 + 1

= − ln∣∣∣∣∣1 +

√1 + u2

u

∣∣∣∣∣+ C = − ln∣∣∣∣∣1 +

√1 + sin2 x

sinx

∣∣∣∣∣+ C

13. u = ex, du = exdx,∫

du√4 + u2

= ln(u+

√u2 + 4

)+ C = ln

(ex +

√e2x + 4

)+ C

14. u = tan−1 x, du =1

1 + x2 dx,

∫eu du = eu + C = etan−1 x + C

15. u =√x− 1, du = 1

2√x− 1

dx, 2∫eu du = 2eu + C = 2e

√x−1 + C

16. u = x2 + 2x, du = (2x+ 2)dx,12

∫cotu du =

12ln | sinu|+ C = 1

2ln sin |x2 + 2x|+ C

17. u =√x, du =

12√xdx,

∫2 coshu du = 2 sinhu+ C = 2 sinh

√x+ C

January 27, 2005 11:45 L24-CH08 Sheet number 2 Page number 338 black

338 Chapter 8

18. u = lnx, du =dx

x,

∫du

u2 = −1u+ C = − 1

lnx+ C

19. u =√x, du =

12√xdx,

∫2 du3u

= 2∫e−u ln 3 du = − 2

ln 3e−u ln 3 + C = − 2

ln 33−√x + C

20. u = sin θ, du = cos θdθ,∫secu tanu du = secu+ C = sec(sin θ) + C

21. u =2x, du = − 2

x2 dx, − 12

∫csch2u du =

12cothu+ C =

12coth

2x+ C

22.∫

dx√x2 − 4

= ln∣∣∣x+√x2 − 4

∣∣∣+ C23. u = e−x, du = −e−xdx, −

∫du

4− u2 = −14ln∣∣∣∣2 + u2− u

∣∣∣∣+ C = −14 ln∣∣∣∣2 + e−x2− e−x

∣∣∣∣+ C24. u = lnx, du =

1xdx,

∫cosu du = sinu+ C = sin(lnx) + C

25. u = ex, du = exdx,∫

ex dx√1− e2x

=∫

du√1− u2

= sin−1 u+ C = sin−1 ex + C

26. u = x−1/2, du = − 12x3/2 dx, −

∫2 sinhu du = −2 coshu+ C = −2 cosh(x−1/2) + C

27. u = x2, du = 2xdx,12

∫du

cscu=12

∫sinu du = −1

2cosu+ C = −1

2cos(x2) + C

28. 2u = ex, 2du = exdx,∫

2du√4− 4u2

= sin−1 u+ C = sin−1(ex/2) + C

29. 4−x2= e−x

2 ln 4, u = −x2 ln 4, du = −2x ln 4 dx = −x ln 16 dx,

− 1ln 16

∫eu du = − 1

ln 16eu + C = − 1

ln 16e−x

2 ln 4 + C = − 1ln 16

4−x2+ C

30. 2πx = eπx ln 2,

∫2πx dx =

1π ln 2

eπx ln 2 + C =1

π ln 22πx + C

31. (a) u = sinx, du = cosx dx,∫u du =

12u2 + C =

12sin2 x+ C

(b)∫sinx cosx dx =

12

∫sin 2x dx = −1

4cos 2x+ C = −1

4(cos2 x− sin2 x) + C

(c) −14(cos2 x− sin2 x) + C = −1

4(1− sin2 x− sin2 x) + C = −1

4+12sin2 x+ C,

and this is the same as the answer in part (a) except for the constants.

32. (a) sech 2x=1

cosh 2x=

1cosh2 x+ sinh2 x

(now multiply top and bottom by sech2x)

=sech2x

1 + tanh2 x

January 27, 2005 11:45 L24-CH08 Sheet number 3 Page number 339 black

Exercise Set 8.2 339

(b)∫sech2x dx =

∫sech2x

1 + tanh2 xdx = tan−1(tanhx) + C, or, replacing 2x with x,

∫sechx dx = tan−1(tanh(x/2)) + C

(c) sechx =1

coshx=

2ex + e−x

=2ex

e2x + 1

(d)∫sechx dx = 2

∫ex

e2x + 1dx = 2 tan−1(ex) + C

33. (a)sec2 xtanx

=1

cos2 x tanx=

1cosx sinx

(b) csc 2x =1

sin 2x=

12 sinx cosx

=12sec2 xtanx

, so∫csc 2x dx =

12ln tanx+ C

(c) secx =1

cosx=

1sin(π/2− x) = csc(π/2− x), so∫

secx dx =∫csc(π/2− x) dx = −1

2ln tan(π/2− x) + C

EXERCISE SET 8.2

1. u = x, dv = e−2xdx, du = dx, v = − 12e−2x;∫

xe−2xdx = −12xe−2x +

∫12e−2xdx = −1

2xe−2x − 1

4e−2x + C

2. u = x, dv = e3xdx, du = dx, v =13e3x;

∫xe3xdx =

13xe3x − 1

3

∫e3xdx =

13xe3x − 1

9e3x + C

3. u = x2, dv = exdx, du = 2x dx, v = ex;∫x2exdx = x2ex − 2

∫xexdx.

For∫xexdx use u = x, dv = exdx, du = dx, v = ex to get

∫xexdx = xex − ex + C1 so

∫x2exdx = x2ex − 2xex + 2ex + C

4. u = x2, dv = e−2xdx, du = 2x dx, v = −12e−2x;

∫x2e−2xdx = −1

2x2e−2x +

∫xe−2xdx

For∫xe−2xdx use u = x, dv = e−2xdx to get

∫xe−2xdx = −1

2xe−2x +

12

∫e−2xdx = −1

2xe−2x − 1

4e−2x + C

so∫x2e−2xdx = −1

2x2e−2x − 1

2xe−2x − 1

4e−2x + C

5. u = x, dv = sin 3x dx, du = dx, v = −13cos 3x;∫

x sin 3x dx = −13x cos 3x+

13

∫cos 3x dx = −1

3x cos 3x+

19sin 3x+ C

January 27, 2005 11:45 L24-CH08 Sheet number 4 Page number 340 black

340 Chapter 8

6. u = x, dv = cos 2x dx, du = dx, v =12sin 2x;∫

x cos 2x dx =12x sin 2x− 1

2

∫sin 2x dx =

12x sin 2x+

14cos 2x+ C

7. u = x2, dv = cosx dx, du = 2x dx, v = sinx;∫x2 cosx dx = x2 sinx− 2

∫x sinx dx

For∫x sinx dx use u = x, dv = sinx dx to get

∫x sinx dx = −x cosx+ sinx+ C1 so

∫x2 cosx dx = x2 sinx+ 2x cosx− 2 sinx+ C

8. u = x2, dv = sinx dx, du = 2x dx, v = − cosx;∫x2 sinx dx = −x2 cosx+ 2

∫x cosx dx; for

∫x cosx dx use u = x, dv = cosx dx to get

∫x cosx dx = x sinx+ cosx+ C1 so

∫x2 sinx dx = −x2 cosx+ 2x sinx+ 2 cosx+ C

9. u = lnx, dv = x dx, du =1xdx, v =

12x2;

∫x lnx dx =

12x2 lnx− 1

2

∫x dx =

12x2 lnx− 1

4x2 + C

10. u = lnx, dv =√x dx, du =

1xdx, v =

23x3/2;∫ √

x lnx dx =23x3/2 lnx− 2

3

∫x1/2dx =

23x3/2 lnx− 4

9x3/2 + C

11. u = (lnx)2, dv = dx, du = 2lnxxdx, v = x;

∫(lnx)2dx = x(lnx)2 − 2

∫lnx dx.

Use u = lnx, dv = dx to get∫lnx dx = x lnx−

∫dx = x lnx− x+ C1 so∫

(lnx)2dx = x(lnx)2 − 2x lnx+ 2x+ C

12. u = lnx, dv =1√xdx, du =

1xdx, v = 2

√x;∫lnx√xdx = 2

√x lnx−2

∫1√xdx = 2

√x lnx−4

√x+C

13. u = ln(3x− 2), dv = dx, du = 33x− 2dx, v = x;

∫ln(3x− 2)dx = x ln(3x− 2)−

∫3x

3x− 2dx

but∫

3x3x− 2dx =

∫ (1 +

23x− 2

)dx = x+

23ln(3x− 2) + C1 so

∫ln(3x− 2)dx = x ln(3x− 2)− x− 2

3ln(3x− 2) + C

14. u = ln(x2 + 4), dv = dx, du =2x

x2 + 4dx, v = x;

∫ln(x2 + 4)dx = x ln(x2 + 4)− 2

∫x2

x2 + 4dx

but∫

x2

x2 + 4dx =

∫ (1− 4

x2 + 4

)dx = x− 2 tan−1 x

2+ C1 so

∫ln(x2 + 4)dx = x ln(x2 + 4)− 2x+ 4 tan−1 x

2+ C

January 27, 2005 11:45 L24-CH08 Sheet number 5 Page number 341 black

Exercise Set 8.2 341

15. u = sin−1 x, dv = dx, du = 1/√1− x2dx, v = x;∫

sin−1 x dx = x sin−1 x−∫x/√1− x2dx = x sin−1 x+

√1− x2 + C

16. u = cos−1(2x), dv = dx, du = − 2√1− 4x2

dx, v = x;∫cos−1(2x)dx = x cos−1(2x) +

∫2x√1− 4x2

dx = x cos−1(2x)− 12

√1− 4x2 + C

17. u = tan−1(3x), dv = dx, du =3

1 + 9x2 dx, v = x;∫tan−1(3x)dx = x tan−1(3x)−

∫3x

1 + 9x2 dx = x tan−1(3x)− 1

6ln(1 + 9x2) + C

18. u = tan−1 x, dv = x dx, du =1

1 + x2 dx, v =12x2;

∫x tan−1 x dx =

12x2 tan−1 x− 1

2

∫x2

1 + x2 dx

but∫

x2

1 + x2 dx =∫ (

1− 11 + x2

)dx = x− tan−1 x+ C1 so∫

x tan−1 x dx =12x2 tan−1 x− 1

2x+

12tan−1 x+ C

19. u = ex, dv = sinx dx, du = exdx, v = − cosx;∫ex sinx dx = −ex cosx+

∫ex cosx dx.

For∫ex cosx dx use u = ex, dv = cosx dx to get

∫ex cosx = ex sinx−

∫ex sinx dx so∫

ex sinx dx = −ex cosx+ ex sinx−∫ex sinx dx,

2∫ex sinx dx = ex(sinx− cosx) + C1,

∫ex sinx dx =

12ex(sinx− cosx) + C

20. u = e3x, dv = cos 2x dx, du = 3e3xdx, v =12sin 2x;∫

e3x cos 2x dx =12e3x sin 2x− 3

2

∫e3x sin 2x dx. Use u = e3x, dv = sin 2x dx to get∫

e3x sin 2x dx = −12e3x cos 2x+

32

∫e3x cos 2x dx, so∫

e3x cos 2x dx =12e3x sin 2x+

34e3x cos 2x− 9

4

∫e3x cos 2x dx,

134

∫e3x cos 2x dx =

14e3x(2 sin 2x+3 cos 2x)+C1,

∫e3x cos 2x dx =

113e3x(2 sin 2x+3 cos 2x)+C

21. u = eax, dv = sin bx dx, du = aeaxdx, v = −1bcos bx (b �= 0);∫

eax sin bx dx = −1beax cos bx+

a

b

∫eax cos bx dx. Use u = eax, dv = cos bx dx to get∫

eax cos bx dx =1beax sin bx− a

b

∫eax sin bx dx so∫

eax sin bx dx = −1beax cos bx+

a

b2eax sin bx− a

2

b2

∫eax sin bx dx,∫

eax sin bx dx =eax

a2 + b2(a sin bx− b cos bx) + C

January 27, 2005 11:45 L24-CH08 Sheet number 6 Page number 342 black

342 Chapter 8

22. From Exercise 21 with a = −3, b = 5, x = θ, answer = e−3θ√34(−3 sin 5θ − 5 cos 5θ) + C

23. u = sin(lnx), dv = dx, du =cos(lnx)

xdx, v = x;∫

sin(lnx)dx = x sin(lnx)−∫cos(lnx)dx. Use u = cos(lnx), dv = dx to get

∫cos(lnx)dx = x cos(lnx) +

∫sin(lnx)dx so

∫sin(lnx)dx = x sin(lnx)− x cos(lnx)−

∫sin(lnx)dx,

∫sin(lnx)dx =

12x[sin(lnx)− cos(lnx)] + C

24. u = cos(lnx), dv = dx, du = − 1xsin(lnx)dx, v = x;∫

cos(lnx)dx = x cos(lnx) +∫sin(lnx)dx. Use u = sin(lnx), dv = dx to get

∫sin(lnx)dx = x sin(lnx)−

∫cos(lnx)dx so

∫cos(lnx)dx = x cos(lnx) + x sin(lnx)−

∫cos(lnx)dx,

∫cos(lnx)dx =

12x[cos(lnx) + sin(lnx)] + C

25. u = x, dv = sec2 x dx, du = dx, v = tanx;∫x sec2 x dx = x tanx−

∫tanx dx = x tanx−

∫sinxcosx

dx = x tanx+ ln | cosx|+ C

26. u = x, dv = tan2 x dx = (sec2 x− 1)dx, du = dx, v = tanx− x;∫x tan2 x dx= x tanx− x2 −

∫(tanx− x)dx

= x tanx− x2 + ln | cosx|+ 12x2 + C = x tanx− 1

2x2 + ln | cosx|+ C

27. u = x2, dv = xex2dx, du = 2x dx, v =

12ex

2;∫

x3ex2dx =

12x2ex

2 −∫xex

2dx =

12x2ex

2 − 12ex

2+ C

28. u = xex, dv =1

(x+ 1)2dx, du = (x+ 1)ex dx, v = − 1

x+ 1;

∫xex

(x+ 1)2dx = − xex

x+ 1+∫exdx = − xex

x+ 1+ ex + C =

ex

x+ 1+ C

29. u = x, dv = e2xdx, du = dx, v =12e2x;

∫ 2

0xe2xdx =

12xe2x

]2

0− 12

∫ 2

0e2xdx = e4 − 1

4e2x]2

0= e4 − 1

4(e4 − 1) = (3e4 + 1)/4

January 27, 2005 11:45 L24-CH08 Sheet number 7 Page number 343 black

Exercise Set 8.2 343

30. u = x, dv = e−5xdx, du = dx, v = −15e−5x;

∫ 1

0xe−5xdx= −1

5xe−5x

]1

0+15

∫ 1

0e−5xdx

= −15e−5 − 1

25e−5x

]1

0= −1

5e−5 − 1

25(e−5 − 1) = (1− 6e−5)/25

31. u = lnx, dv = x2dx, du =1xdx, v =

13x3;

∫ e

1x2 lnx dx =

13x3 lnx

]e1− 13

∫ e

1x2dx =

13e3 − 1

9x3]e

1=13e3 − 1

9(e3 − 1) = (2e3 + 1)/9

32. u = lnx, dv =1x2 dx, du =

1xdx, v = − 1

x;

∫ e

√e

lnxx2 dx= −

1xlnx]e√e

+∫ e

√e

1x2 dx

= −1e+1√eln√e− 1

x

]e√e

= −1e+

12√e− 1e+1√e=3√e− 42e

33. u = ln(x+ 2), dv = dx, du =1

x+ 2dx, v = x;

∫ 1

−1ln(x+ 2)dx= x ln(x+ 2)

]1

−1−∫ 1

−1

x

x+ 2dx = ln 3 + ln 1−

∫ 1

−1

[1− 2

x+ 2

]dx

= ln 3− [x− 2 ln(x+ 2)]]1

−1= ln 3− (1− 2 ln 3) + (−1− 2 ln 1) = 3 ln 3− 2

34. u = sin−1 x, dv = dx, du =1√1− x2

dx, v = x;

∫ √3/2

0sin−1 x dx= x sin−1 x

]√3/2

0−∫ √3/2

0

x√1− x2

dx =√32sin−1

√32+√1− x2

]√3/2

0

=√32

(π3

)+12− 1 = π

√36− 12

35. u = sec−1√θ, dv = dθ, du =

12θ√θ − 1

dθ, v = θ;

∫ 4

2sec−1

√θdθ= θ sec−1

√θ

]4

2− 12

∫ 4

2

1√θ − 1

dθ = 4 sec−1 2− 2 sec−1√2−√θ − 1

]4

2

= 4(π3

)− 2

(π4

)−√3 + 1 =

5π6−√3 + 1

36. u = sec−1 x, dv = x dx, du =1

x√x2 − 1

dx, v =12x2;

∫ 2

1x sec−1 x dx=

12x2 sec−1 x

]2

1− 12

∫ 2

1

x√x2 − 1

dx

=12[(4)(π/3)− (1)(0)]− 1

2

√x2 − 1

]2

1= 2π/3−

√3/2

January 27, 2005 11:45 L24-CH08 Sheet number 8 Page number 344 black

344 Chapter 8

37. u = x, dv = sin 2x dx, du = dx, v = −12cos 2x;

∫ π

0x sin 2x dx = −1

2x cos 2x

]π0+12

∫ π

0cos 2x dx = −π/2 + 1

4sin 2x

]π0= −π/2

38.∫ π

0(x+ x cosx)dx =

12x2]π

0+∫ π

0x cosx dx =

π2

2+∫ π

0x cosx dx;

u = x, dv = cosx dx, du = dx, v = sinx∫ π

0x cosx dx = x sinx

]π0−∫ π

0sinx dx = cosx

]π0= −2 so

∫ π

0(x+ x cosx)dx = π2/2− 2

39. u = tan−1√x, dv =√xdx, du =

12√x(1 + x)

dx, v =23x3/2;

∫ 3

1

√x tan−1√xdx= 2

3x3/2 tan−1√x

]3

1− 13

∫ 3

1

x

1 + xdx

=23x3/2 tan−1√x

]3

1− 13

∫ 3

1

[1− 1

1 + x

]dx

=[23x3/2 tan−1√x− 1

3x+

13ln |1 + x|

]3

1= (2√3π − π/2− 2 + ln 2)/3

40. u = ln(x2 + 1), dv = dx, du =2x

x2 + 1dx, v = x;

∫ 2

0ln(x2 + 1)dx= x ln(x2 + 1)

]2

0−∫ 2

0

2x2

x2 + 1dx = 2 ln 5− 2

∫ 2

0

(1− 1

x2 + 1

)dx

= 2 ln 5− 2(x− tan−1 x)]2

0= 2 ln 5− 4 + 2 tan−1 2

41. t =√x, t2 = x, dx = 2t dt

(a)∫e√xdx = 2

∫tet dt; u = t, dv = etdt, du = dt, v = et,

∫e√xdx = 2tet − 2

∫et dt = 2(t− 1)et + C = 2(

√x− 1)e

√x + C

(b)∫cos√x dx = 2

∫t cos t dt; u = t, dv = cos tdt, du = dt, v = sin t,

∫cos√x dx = 2t sin t− 2

∫sin tdt = 2t sin t+ 2 cos t+ C = 2

√x sin

√x+ 2 cos

√x+ C

January 27, 2005 11:45 L24-CH08 Sheet number 9 Page number 345 black

Exercise Set 8.2 345

42. Let q1(x), q2(x), q3(x) denote successive antiderivatives of q(x),so that q′3(x) = q2(x), q

′2(x) = q1(x), q

′1(x) = q(x). Let p(x) = ax

2 + bx+ c.

Repeated RepeatedDifferentiation Antidifferentiation

ax2 + bx+ c q(x)+

2ax+ b q1(x)−

2a q2(x)+

0 q3(x)

Then∫p(x)q(x) dx = (ax2 + bx+ c)q1(x)− (2ax+ b)q2(x) + 2aq3(x) + C. Check:

d

dx[(ax2+bx+ c)q1(x)− (2ax+ b)q2(x) + 2aq3(x)]

= (2ax+ b)q1(x) + (ax2 + bx+ c)q(x)− 2aq2(x)− (2ax+ b)q1(x) + 2aq2(x) = p(x)q(x)

43. Repeated RepeatedDifferentiation Antidifferentiation

3x2 − x+ 2 e−x

+6x− 1 −e−x

−6 e−x

+0 −e−x∫

(3x2 − x+ 2)e−x = −(3x2 − x+ 2)e−x − (6x− 1)e−x − 6e−x + C = −e−x[3x2 + 5x+ 7] + C

44. Repeated RepeatedDifferentiation Antidifferentiation

x2 + x+ 1 sinx+

2x+ 1 − cosx−

2 − sinx+

0 cosx∫(x2 + x+ 1) sinx dx= −(x2 + x+ 1) cosx+ (2x+ 1) sinx+ 2 cosx+ C

= −(x2 + x− 1) cosx+ (2x+ 1) sinx+ C

January 27, 2005 11:45 L24-CH08 Sheet number 10 Page number 346 black

346 Chapter 8

45. Repeated RepeatedDifferentiation Antidifferentiation

4x4 sin 2x+

16x3 −12cos 2x

−48x2 −1

4sin 2x

+

96x18cos 2x

96116sin 2x

+

0 − 132cos 2x

∫4x4 sin 2x dx = (−2x4 + 6x2 − 3) cos 2x+−(4x3 + 6x) sin 2x+ C

46. Repeated RepeatedDifferentiation Antidifferentiation

x3 √2x+ 1

+

3x2 13(2x+ 1)3/2

−6x

115(2x+ 1)5/2

+

61105

(2x+ 1)7/2

−0

1945

(2x+ 1)9/2

∫x3√2x+ 1 dx = 1

3x3(2x+ 1)3/2 − 1

5x2(2x+ 1)5/2 +

235x(2x+ 1)7/2 − 2

315(2x+ 1)9/2 + C

47. (a) We perform a single integration by parts:

u = cosx, dv = sinx dx, du = − sinx dx, v = − cosx,∫sinx cosx dx = − cos2 x−

∫sinx cosx dx. Thus

2∫sinx cosx dx = − cos2 x+ C,

∫sinx cosx dx = −1

2cos2 x+ C

(b) u = sinx, du = cosx dx,∫sinx cosx dx =

∫u du =

12u2 + C =

12sin2 x+ C

48. (a) u = x2, dv =x√x2 + 1

, du = 2x dx, v =√x2 + 1,

∫ 1

0

x3√x2 + 1

dx = x2√x2 + 1

]1

0−∫ 1

02x√x2 + 1 dx =

√2− 2

3(x2 + 1)3/2

]1

0= −1

3

√2 +

23

January 27, 2005 11:45 L24-CH08 Sheet number 11 Page number 347 black

Exercise Set 8.2 347

(b) u =√x2 + 1, du =

x√x2 + 1

dx,

∫ √2

1(u2 − 1) du =

(13u3 − u

)]√2

1

=23

√2−√2− 1

3+ 1 = −1

3

√2 +

23.

49. (a) A =∫ e

1lnx dx = (x lnx− x)

]e1= 1

(b) V = π∫ e

1(lnx)2dx = π

[(x(lnx)2 − 2x lnx+ 2x)

]e1= π(e− 2)

50. A =∫ π/2

0(x− x sinx)dx = 1

2x2]π/2

0−∫ π/2

0x sinx dx =

π2

8− (−x cosx+ sinx)

]π/20

= π2/8− 1

51. V = 2π∫ π

0x sinx dx = 2π(−x cosx+ sinx)

]π0= 2π2

52. V = 2π∫ π/2

0x cosx dx = 2π(cosx+ x sinx)

]π/20

= π(π − 2)

53. distance =∫ π

0t3 sin tdt;

Repeated RepeatedDifferentiation Antidifferentiation

t3 sin t+

3t2 − cos t−

6t − sin t+

6 cos t−

0 sin t

∫ π

0t3 sin t dx = [(−t3 cos t+ 3t2 sin t+ 6t cos t− 6 sin t)]

]π0= π3 − 6π

54. u = 2t, dv = sin(kωt)dt, du = 2dt, v = − 1kωcos(kωt); the integrand is an even function of t so

∫ π/ω

−π/ωt sin(kωt) dt = 2

∫ π/ω

0t sin(kωt) dt = − 2

kωt cos(kωt)

]π/ω0

+ 2∫ π/ω

0

1kωcos(kωt) dt

=2π(−1)k+1

kω2 +2

k2ω2 sin(kωt)]π/ω

0=2π(−1)k+1

kω2

55. (a)∫sin4 x dx= −1

4sin3 x cosx+

34

∫sin2 x dx

= −14sin3 x cosx+

34

[−12sinx cosx+

12x

]+ C

= −14sin3 x cosx− 3

8sinx cosx+

38x+ C

January 27, 2005 11:45 L24-CH08 Sheet number 12 Page number 348 black

348 Chapter 8

(b)∫ π/2

0sin5 x dx= −1

5sin4 x cosx

]π/20

+45

∫ π/2

0sin3 x dx

=45

[−13sin2 x cosx

]π/20

+23

∫ π/2

0sinx dx

]

= − 815cosx

]π/20

=815

56. (a)∫cos5 x dx=

15cos4 x sinx+

45

∫cos3 x dx =

15cos4 x sinx+

45

[13cos2 x sinx+

23sinx

]+ C

=15cos4 x sinx+

415cos2 x sinx+

815sinx+ C

(b)∫cos6 x dx=

16cos5 x sinx+

56

∫cos4 x dx

=16cos5 x sinx+

56

[14cos3 x sinx+

34

∫cos2 x dx

]

=16cos5 x sinx+

524cos3 x sinx+

58

[12cosx sinx+

12x

]+ C,

[16cos5 x sinx+

524cos3 x sinx+

516cosx sinx+

516x

]π/20

= 5π/32

57. u = sinn−1 x, dv = sinx dx, du = (n− 1) sinn−2 x cosx dx, v = − cosx;∫sinn x dx= − sinn−1 x cosx+ (n− 1)

∫sinn−2 x cos2 x dx

= − sinn−1 x cosx+ (n− 1)∫sinn−2 x (1− sin2 x)dx

= − sinn−1 x cosx+ (n− 1)∫sinn−2 x dx− (n− 1)

∫sinn x dx,

n

∫sinn x dx = − sinn−1 x cosx+ (n− 1)

∫sinn−2 x dx,

∫sinn x dx = − 1

nsinn−1 x cosx+

n− 1n

∫sinn−2 x dx

58. (a) u = secn−2 x, dv = sec2 x dx, du = (n− 2) secn−2 x tanx dx, v = tanx;∫secn x dx= secn−2 x tanx− (n− 2)

∫secn−2 x tan2 x dx

= secn−2 x tanx− (n− 2)∫secn−2 x (sec2 x− 1)dx

= secn−2 x tanx− (n− 2)∫secn x dx+ (n− 2)

∫secn−2 x dx,

(n− 1)∫secn x dx = secn−2 x tanx+ (n− 2)

∫secn−2 x dx,

∫secn x dx =

1n− 1 sec

n−2 x tanx+n− 2n− 1

∫secn−2 x dx

January 27, 2005 11:45 L24-CH08 Sheet number 13 Page number 349 black

Exercise Set 8.2 349

(b)∫tann x dx =

∫tann−2 x (sec2 x− 1) dx =

∫tann−1 x sec2 x dx−

∫tann−2 x dx

=1

n− 1 tann−1 x−

∫tann−2 x dx

(c) u = xn, dv = exdx, du = nxn−1dx, v = ex;∫xnexdx = xnex − n

∫xn−1exdx

59. (a)∫tan4 x dx =

13tan3 x−

∫tan2 x dx =

13tan3 x− tanx+

∫dx =

13tan3 x− tanx+ x+ C

(b)∫sec4 x dx =

13sec2 x tanx+

23

∫sec2 x dx =

13sec2 x tanx+

23tanx+ C

(c)∫x3exdx= x3ex − 3

∫x2exdx = x3ex − 3

[x2ex − 2

∫xexdx

]

= x3ex − 3x2ex + 6[xex −

∫exdx

]= x3ex − 3x2ex + 6xex − 6ex + C

60. (a) u = 3x,∫x2e3xdx=

127

∫u2eudu =

127

[u2eu − 2

∫ueudu

]=127u2eu − 2

27

[ueu −

∫eudu

]

=127u2eu − 2

27ueu +

227eu + C =

13x2e3x − 2

9xe3x +

227e3x + C

(b) u = −√x,

∫ 1

0xe−

√xdx = 2

∫ −1

0u3eudu,

∫u3eudu= u3eu − 3

∫u2eudu = u3eu − 3

[u2eu − 2

∫ueudu

]

= u3eu − 3u2eu + 6[ueu −

∫eudu

]= u3eu − 3u2eu + 6ueu − 6eu + C,

2∫ −1

0u3eudu = 2(u3 − 3u2 + 6u− 6)eu

]−1

0= 12− 32e−1

61. u = x, dv = f ′′(x)dx, du = dx, v = f ′(x);∫ 1

−1x f ′′(x)dx = xf ′(x)

]1

−1−∫ 1

−1f ′(x)dx

= f ′(1) + f ′(−1)− f(x)]1

−1= f ′(1) + f ′(−1)− f(1) + f(−1)

62. (a)∫u dv = uv −

∫v du = x(sinx+ C1) + cosx− C1x+ C2 = x sinx+ cosx+ C2;

the constant C1 cancels out and hence plays no role in the answer.

(b) u(v + C1)−∫(v + C1)du = uv + C1u−

∫v du− C1u = uv −

∫v du

January 27, 2005 11:45 L24-CH08 Sheet number 14 Page number 350 black

350 Chapter 8

63. u = ln(x+ 1), dv = dx, du =dx

x+ 1, v = x+ 1;∫

ln(x+ 1) dx =∫u dv = uv −

∫v du = (x+ 1) ln(x+ 1)−

∫dx = (x+ 1) ln(x+ 1)− x+ C

64. u = ln(3x− 2), dv = dx, du = 3dx3x− 2 , v = x−

23;

∫ln(3x− 2) dx=

∫u dv = uv −

∫v du =

(x− 2

3

)ln(3x− 2)−

∫ (x− 2

3

)1

x− 2/3 dx

=(x− 2

3

)ln(3x− 2)−

(x− 2

3

)+ C

65. u = tan−1 x, dv = x dx, du =1

1 + x2 dx, v =12(x2 + 1)∫

x tan−1 x dx=∫u dv = uv −

∫v du =

12(x2 + 1) tan−1 x− 1

2

∫dx

=12(x2 + 1) tan−1 x− 1

2x+ C

66. u =1lnx

, dv = 1x dx, du = − 1

x(ln x)2 dx, v = lnx∫1

x lnxdx = 1 +

∫1

x lnxdx.

This seems to imply that 1 = 0, but recall that both sides represent a function plus an arbitraryconstant; these two arbitrary constants will take care of the 1.

67. (a) u = f(x), dv = dx, du = f ′(x), v = x;∫ b

a

f(x) dx = xf(x)]ba

−∫ b

a

xf ′(x) dx = bf(b)− af(a)−∫ b

a

xf ′(x) dx

(b) Substitute y = f(x), dy = f ′(x) dx, x = a when y = f(a), x = b when y = f(b),∫ b

a

xf ′(x) dx =∫ f(b)

f(a)x dy =

∫ f(b)

f(a)f−1(y) dy

(c) From a = f−1(α) and b = f−1(β) we get

bf(b)− af(a) = βf−1(β)− αf−1(α); then∫ β

α

f−1(x) dx =∫ β

α

f−1(y) dy =∫ f(b)

f(a)f−1(y) dy,

which, by Part (b), yields

a

b

a = f –1(a) b = f –1(b)

x

y

A1

A2

∫ β

α

f−1(x) dx = bf(b)− af(a)−∫ b

a

f(x) dx

= βf−1(β)− αf−1(α)−∫ f−1(β)

f−1(α)f(x) dx

Note from the figure that A1 =∫ β

α

f−1(x) dx,A2 =∫ f−1(β)

f−1(α)f(x) dx, and

A1 +A2 = βf−1(β)− αf−1(α), a “picture proof”.

January 27, 2005 11:45 L24-CH08 Sheet number 15 Page number 351 black

Exercise Set 8.3 351

68. (a) Use Exercise 67(c);∫ 1/2

0sin−1 x dx =

12sin−1

(12

)−0·sin−1 0−

∫ sin−1(1/2)

sin−1(0)sinx dx =

12sin−1

(12

)−∫ π/6

0sinx dx

(b) Use Exercise 67(b);∫ e2

e

lnx dx = e2 ln e2 − e ln e−∫ ln e2

ln ef−1(y) dy = 2e2 − e−

∫ 2

1ey dy = 2e2 − e−

∫ 2

1ex dx

EXERCISE SET 8.3

1. u = cosx, −∫u3du = −1

4cos4 x+ C

2. u = sin 3x,13

∫u5 du =

118sin6 3x+ C

3.∫sin2 5θ =

12

∫(1− cos 10θ) dθ = 1

2θ − 1

20sin 10θ + C

4.∫cos2 3x dx =

12

∫(1 + cos 6x)dx =

12x+

112sin 6x+ C

5.∫sin3 aθ dθ =

∫sin aθ(1− cos2 aθ) dθ = −1

acos aθ − 1

3acos3 aθ + C (a �= 0)

6.∫cos3 at dt=

∫(1− sin2 at) cos at dt

=∫cos at dt−

∫sin2 at cos at dt =

1asin at− 1

3asin3 at+ C (a �= 0)

7. u = sin ax,1a

∫u du =

12asin2 ax+ C, a �= 0

8.∫sin3 x cos3 x dx=

∫sin3 x(1− sin2 x) cosx dx

=∫(sin3 x− sin5 x) cosx dx =

14sin4 x− 1

6sin6 x+ C

9.∫sin2 t cos3 t dt=

∫sin2 t(1− sin2 t) cos t dt =

∫(sin2 t− sin4 t) cos t dt

=13sin3 t− 1

5sin5 t+ C

10.∫sin3 x cos2 x dx=

∫(1− cos2 x) cos2 x sinx dx

=∫(cos2 x− cos4 x) sinx dx = −1

3cos3 x+

15cos5 x+ C

11.∫sin2 x cos2 x dx =

14

∫sin2 2x dx =

18

∫(1− cos 4x)dx = 1

8x− 1

32sin 4x+ C

January 27, 2005 11:45 L24-CH08 Sheet number 16 Page number 352 black

352 Chapter 8

12.∫sin2 x cos4 x dx=

18

∫(1− cos 2x)(1 + cos 2x)2dx = 1

8

∫(1− cos2 2x)(1 + cos 2x)dx

=18

∫sin2 2x dx+

18

∫sin2 2x cos 2x dx =

116

∫(1− cos 4x)dx+ 1

48sin3 2x

=116x− 1

64sin 4x+

148sin3 2x+ C

13.∫sin 2x cos 3x dx =

12

∫(sin 5x− sinx)dx = − 1

10cos 5x+

12cosx+ C

14.∫sin 3θ cos 2θdθ =

12

∫(sin 5θ + sin θ)dθ = − 1

10cos 5θ − 1

2cos θ + C

15.∫sinx cos(x/2)dx =

12

∫[sin(3x/2) + sin(x/2)]dx = −1

3cos(3x/2)− cos(x/2) + C

16. u = cosx, −∫u1/3du = −3

4cos4/3 x+ C

17.∫ π/2

0cos3 x dx=

∫ π/2

0(1− sin2 x) cosx dx

=[sinx− 1

3sin3 x

]π/20

=23

18.∫ π/2

0sin2(x/2) cos2(x/2)dx=

14

∫ π/2

0sin2 x dx =

18

∫ π/2

0(1− cos 2x)dx

=18

(x− 1

2sin 2x

)]π/20

= π/16

19.∫ π/3

0sin4 3x cos3 3x dx =

∫ π/3

0sin4 3x(1− sin2 3x) cos 3x dx =

[115sin5 3x− 1

21sin7 3x

]π/30= 0

20.∫ π

−πcos2 5θ dθ =

12

∫ π

−π(1 + cos 10θ)dθ =

12

(θ +

110sin 10θ

)]π−π= π

21.∫ π/6

0sin 4x cos 2x dx=

12

∫ π/6

0(sin 2x+ sin 6x)dx =

[−14cos 2x− 1

12cos 6x

]π/60

= [(−1/4)(1/2)− (1/12)(−1)]− [−1/4− 1/12] = 7/24

22.∫ 2π

0sin2 kx dx =

12

∫ 2π

0(1− cos 2kx)dx = 1

2

(x− 1

2ksin 2kx

)]2π

0= π − 1

4ksin 4πk (k �= 0)

23.12tan(2x− 1) + C 24. −1

5ln | cos 5x|+ C

25. u = e−x, du = −e−x dx; −∫tanu du = ln | cosu|+ C = ln | cos(e−x)|+ C

26.13ln | sin 3x|+ C 27.

14ln | sec 4x+ tan 4x|+ C

January 27, 2005 11:45 L24-CH08 Sheet number 17 Page number 353 black

Exercise Set 8.3 353

28. u =√x, du =

12√xdx;

∫2 secu du = 2 ln | secu+ tanu|+ C = 2 ln

∣∣sec√x+ tan√x∣∣+ C29. u = tanx,

∫u2du =

13tan3 x+ C

30.∫tan5 x(1 + tan2 x) sec2 x dx =

∫(tan5 x+ tan7 x) sec2 x dx =

16tan6 x+

18tan8 x+ C

31.∫tan 4x(1 + tan2 4x) sec2 4x dx =

∫(tan 4x+ tan3 4x) sec2 4x dx =

18tan2 4x+

116tan4 4x+ C

32.∫tan4 θ(1 + tan2 θ) sec2 θ dθ =

15tan5 θ +

17tan7 θ + C

33.∫sec4 x(sec2 x− 1) secx tanx dx =

∫(sec6 x− sec4 x) secx tanx dx = 1

7sec7 x− 1

5sec5 x+ C

34.∫(sec2 θ− 1)2 sec θ tan θdθ =

∫(sec4 θ− 2 sec2 θ+1) sec θ tan θdθ = 1

5sec5 θ− 2

3sec3 θ+ sec θ+C

35.∫(sec2 x− 1)2 secx dx =

∫(sec5 x− 2 sec3 x+ secx)dx =

∫sec5 x dx− 2

∫sec3 x dx+

∫secx dx

=14sec3 x tanx+

34

∫sec3 x dx− 2

∫sec3 x dx+ ln | secx+ tanx|

=14sec3 x tanx− 5

4

[12secx tanx+

12ln | secx+ tanx|

]+ ln | secx+ tanx|+ C

=14sec3 x tanx− 5

8secx tanx+

38ln | secx+ tanx|+ C

36.∫[sec2 x− 1] sec3 x dx =

∫[sec5 x− sec3 x]dx

=(14sec3 x tanx+

34

∫sec3 x dx

)−∫sec3 x dx (equation (20))

=14sec3 x tanx− 1

4

∫sec3 x dx

=14sec3 x tanx− 1

8secx tanx− 1

8ln | secx+ tanx|+ C (equation (20), (22))

37.∫sec2 t(sec t tan t)dt =

13sec3 t+ C 38.

∫sec4 x(secx tanx)dx =

15sec5 x+ C

39.∫sec4 x dx =

∫(1 + tan2 x) sec2 x dx =

∫(sec2 x+ tan2 x sec2 x)dx = tanx+

13tan3 x+ C

40. Using equation (20),∫sec5 x dx=

14sec3 x tanx+

34

∫sec3 x dx

=14sec3 x tanx+

38secx tanx+

38ln | secx+ tanx|+ C

January 27, 2005 11:45 L24-CH08 Sheet number 18 Page number 354 black

354 Chapter 8

41. u = 4x, use equation (19) to get14

∫tan3 u du =

14

[12tan2 u+ ln | cosu|

]+ C =

18tan2 4x+

14ln | cos 4x|+ C

42. Use equation (19) to get∫tan4 x dx =

13tan3 x− tanx+ x+ C

43.∫ √

tanx(1 + tan2 x) sec2 x dx =23tan3/2 x+

27tan7/2 x+ C

44.∫sec1/2 x(secx tanx)dx =

23sec3/2 x+ C

45.∫ π/8

0(sec2 2x− 1)dx =

[12tan 2x− x

]π/80

= 1/2− π/8

46.∫ π/6

0sec2 2θ(sec 2θ tan 2θ)dθ =

16sec3 2θ

]π/60

= (1/6)(2)3 − (1/6)(1) = 7/6

47. u = x/2,

2∫ π/4

0tan5 u du =

[12tan4 u− tan2 u− 2 ln | cosu|

]π/40

= 1/2− 1− 2 ln(1/√2) = −1/2 + ln 2

48. u = πx,1π

∫ π/4

0secu tanu du =

1πsecu

]π/40

= (√2− 1)/π

49.∫(csc2 x− 1) csc2 x(cscx cotx)dx =

∫(csc4 x− csc2 x)(cscx cotx)dx = −1

5csc5 x+

13csc3 x+ C

50.∫cos2 3tsin2 3t

· 1cos 3t

dt =∫csc 3t cot 3t dt = −1

3csc 3t+ C

51.∫(csc2 x− 1) cotx dx =

∫cscx(cscx cotx)dx−

∫cosxsinx

dx = −12csc2 x− ln | sinx|+ C

52.∫(cot2 x+ 1) csc2 x dx = −1

3cot3 x− cotx+ C

53. (a)∫ 2π

0sinmx cosnx dx =

12

∫ 2π

0[sin(m+ n)x+ sin(m− n)x]dx

=[−cos(m+ n)x

2(m+ n)− cos(m− n)x

2(m− n)

]2π

0

but cos(m+ n)x]2π

0= 0, cos(m− n)x

]2π

0= 0.

(b)∫ 2π

0cosmx cosnx dx =

12

∫ 2π

0[cos(m+ n)x+ cos(m− n)x]dx;

since m �= n, evaluate sin at integer multiples of 2π to get 0.

(c)∫ 2π

0sinmx sinnx dx =

12

∫ 2π

0[cos(m− n)x− cos(m+ n)x] dx;

since m �= n, evaluate sin at integer multiples of 2π to get 0.

January 27, 2005 11:45 L24-CH08 Sheet number 19 Page number 355 black

Exercise Set 8.3 355

54. (a)∫ 2π

0sinmx cosmxdx =

12

∫ 2π

0sin 2mxdx = − 1

4mcos 2mx

∣∣∣∣∣2π

0

= 0

(b)∫ 2π

0cos2mxdx =

12

∫ 2π

0(1 + cos 2mx) dx =

12

(x+

12m

sin 2mx) ∣∣∣∣∣

0

= π

(c)∫ 2π

0sin2mxdx =

12

∫ 2π

0(1− cos 2mx) dx = 1

2

(x− 1

2msin 2mx

) ∣∣∣∣∣2π

0

= π

55. y′ = tanx, 1 + (y′)2 = 1 + tan2 x = sec2 x,

L =∫ π/4

0

√sec2 x dx =

∫ π/4

0secx dx = ln | secx+ tanx|

]π/40

= ln(√2 + 1)

56. V = π∫ π/4

0(1− tan2 x)dx = π

∫ π/4

0(2− sec2 x)dx = π(2x− tanx)

]π/40

=12π(π − 2)

57. V = π∫ π/4

0(cos2 x− sin2 x)dx = π

∫ π/4

0cos 2x dx =

12π sin 2x

]π/40

= π/2

58. V = π∫ π

0sin2 x dx =

π

2

∫ π

0(1− cos 2x)dx = π

2

(x− 1

2sin 2x

)]π0= π2/2

59. With 0 < α < β,D = Dβ−Dα =L

∫ β

α

secx dx =L

2πln | secx+ tanx|

]βα

=L

2πln∣∣∣∣ secβ + tanβsecα+ tanα

∣∣∣∣60. (a) D =

1002π

ln(sec 25◦ + tan 25◦) = 7.18 cm

(b) D =1002π

ln∣∣∣∣ sec 50◦ + tan 50◦sec 30◦ + tan 30◦

∣∣∣∣ = 7.34 cm61. (a)

∫cscx dx =

∫sec(π/2− x)dx= − ln | sec(π/2− x) + tan(π/2− x)|+ C

= − ln | cscx+ cotx|+ C

(b) − ln | cscx+ cotx|= ln 1| cscx+ cotx| = ln

| cscx− cotx|| csc2 x− cot2 x| = ln | cscx− cotx|,

− ln | cscx+ cotx|= − ln∣∣∣∣ 1sinx

+cosxsinx

∣∣∣∣ = ln∣∣∣∣ sinx1 + cosx

∣∣∣∣= ln

∣∣∣∣2 sin(x/2) cos(x/2)2 cos2(x/2)

∣∣∣∣ = ln | tan(x/2)|62. sinx+ cosx=

√2[(1/√2) sinx+ (1/

√2) cosx

]=√2 [sinx cos(π/4) + cosx sin(π/4)] =

√2 sin(x+ π/4),

∫dx

sinx+ cosx=

1√2

∫csc(x+ π/4)dx = − 1√

2ln | csc(x+ π/4) + cot(x+ π/4)|+ C

= − 1√2ln

∣∣∣∣∣√2 + cosx− sinxsinx+ cosx

∣∣∣∣∣+ C

January 27, 2005 11:45 L24-CH08 Sheet number 20 Page number 356 black

356 Chapter 8

63. a sinx+ b cosx =√a2 + b2

[a√

a2 + b2sinx+

b√a2 + b2

cosx]=√a2 + b2(sinx cos θ+ cosx sin θ)

where cos θ = a/√a2 + b2 and sin θ = b/

√a2 + b2 so a sinx+ b cosx =

√a2 + b2 sin(x+ θ)

and∫

dx

a sinx+ b cosx=

1√a2 + b2

∫csc(x+ θ)dx = − 1√

a2 + b2ln | csc(x+ θ) + cot(x+ θ)|+ C

= − 1√a2 + b2

ln

∣∣∣∣∣√a2 + b2 + a cosx− b sinx

a sinx+ b cosx

∣∣∣∣∣+ C

64. (a)∫ π/2

0sinn x dx = − 1

nsinn−1 x cosx

]π/20

+n− 1n

∫ π/2

0sinn−2 x dx =

n− 1n

∫ π/2

0sinn−2 x dx

(b) By repeated application of the formula in Part (a)∫ π/2

0sinn x dx=

(n− 1n

)(n− 3n− 2

)∫ π/2

0sinn−4 x dx

=

(n− 1n

)(n− 3n− 2

)(n− 5n− 4

)· · ·(12

)∫ π/2

0dx, n even

(n− 1n

)(n− 3n− 2

)(n− 5n− 4

)· · ·(23

)∫ π/2

0sinx dx, n odd

=

1 · 3 · 5 · · · (n− 1)2 · 4 · 6 · · ·n · π

2, n even

2 · 4 · 6 · · · (n− 1)3 · 5 · 7 · · ·n , n odd

65. (a)∫ π/2

0sin3 x dx =

23

(b)∫ π/2

0sin4 x dx =

1 · 32 · 4 ·

π

2= 3π/16

(c)∫ π/2

0sin5 x dx =

2 · 43 · 5 = 8/15 (d)

∫ π/2

0sin6 x dx =

1 · 3 · 52 · 4 · 6 ·

π

2= 5π/32

66. Similar to proof in Exercise 64.

EXERCISE SET 8.4

1. x = 2 sin θ, dx = 2 cos θ dθ,

4∫cos2 θ dθ= 2

∫(1 + cos 2θ)dθ = 2θ + sin 2θ + C

= 2θ + 2 sin θ cos θ + C = 2 sin−1(x/2) +12x√4− x2 + C

2. x =12sin θ, dx =

12cos θ dθ,

12

∫cos2 θ dθ=

14

∫(1 + cos 2θ)dθ =

14θ +

18sin 2θ + C

=14θ +

14sin θ cos θ + C =

14sin−1 2x+

12x√1− 4x2 + C

January 27, 2005 11:45 L24-CH08 Sheet number 21 Page number 357 black

Exercise Set 8.4 357

3. x = 4 sin θ, dx = 4 cos θ dθ,

16∫sin2 θ dθ= 8

∫(1− cos 2θ)dθ = 8θ − 4 sin 2θ + C = 8θ − 8 sin θ cos θ + C

= 8 sin−1(x/4)− 12x√16− x2 + C

4. x = 3 sin θ, dx = 3 cos θ dθ,

19

∫1

sin2 θdθ =

19

∫csc2 θ dθ = −1

9cot θ + C = −

√9− x2

9x+ C

5. x = 2 tan θ, dx = 2 sec2 θ dθ,18

∫1

sec2 θdθ=

18

∫cos2 θ dθ =

116

∫(1 + cos 2θ)dθ =

116θ +

132sin 2θ + C

=116θ +

116sin θ cos θ + C =

116tan−1 x

2+

x

8(4 + x2)+ C

6. x =√5 tan θ, dx =

√5 sec2 θ dθ,

5∫tan2 θ sec θ dθ = 5

∫(sec3 θ − sec θ)dθ = 5

(12sec θ tan θ − 1

2ln | sec θ + tan θ|

)+ C1

=12x√5 + x2 − 5

2ln√5 + x2 + x√

5+ C1 =

12x√5 + x2 − 5

2ln(√5 + x2 + x) + C

7. x = 3 sec θ, dx = 3 sec θ tan θ dθ,

3∫tan2 θ dθ = 3

∫(sec2 θ − 1)dθ = 3 tan θ − 3θ + C =

√x2 − 9− 3 sec−1 x

3+ C

8. x = 4 sec θ, dx = 4 sec θ tan θ dθ,

116

∫1sec θ

dθ =116

∫cos θ dθ =

116sin θ + C =

√x2 − 1616x

+ C

9. x = sin θ, dx = cos θ dθ,

3∫sin3 θ dθ = 3

∫ [1− cos2 θ

]sin θ dθ

= 3(− cos θ + cos3 θ

)+ C = −3

√1− x2 + (1− x2)3/2 + C

10. x =√5 sin θ, dx =

√5 cos θ dθ,

25√5∫sin3 θ cos2 θ dθ = 25

√5(−13cos3 θ +

15cos5 θ

)+ C = −5

3(5− x2)3/2 +

15(5− x2)5/2 + C

11. x =23sec θ, dx =

23sec θ tan θ dθ,

34

∫1sec θ

dθ =34

∫cos θ dθ =

34sin θ + C =

14x

√9x2 − 4 + C

12. t = tan θ, dt = sec2 θ dθ,∫sec3 θtan θ

dθ=∫tan2 θ + 1tan θ

sec θ dθ =∫(sec θ tan θ + csc θ)dθ

= sec θ + ln | csc θ − cot θ|+ C =√1 + t2 + ln

√1 + t2 − 1|t| + C

13. x = sin θ, dx = cos θ dθ,∫

1cos2 θ

dθ =∫sec2 θ dθ = tan θ + C = x/

√1− x2 + C

January 27, 2005 11:45 L24-CH08 Sheet number 22 Page number 358 black

358 Chapter 8

14. x = 5 tan θ, dx = 5 sec2 θ dθ,125

∫sec θtan2 θ

dθ =125

∫csc θ cot θ dθ = − 1

25csc θ+C = −

√x2 + 2525x

+ C

15. x = 3 sec θ, dx = 3 sec θ tan θ dθ,∫sec θ dθ = ln | sec θ + tan θ|+ C = ln

∣∣∣∣13x+ 13√x2 − 9

∣∣∣∣+ C16. 1 + 2x2 + x4 = (1 + x2)2, x = tan θ, dx = sec2 θ dθ,∫

1sec2 θ

dθ=∫cos2 θ dθ =

12

∫(1 + cos 2θ)dθ =

12θ +

14sin 2θ + C

=12θ +

12sin θ cos θ + C =

12tan−1 x+

x

2(1 + x2)+ C

17. x =32sec θ, dx =

32sec θ tan θ dθ,

32

∫sec θ tan θ dθ27 tan3 θ

=118

∫cos θsin2 θ

dθ = − 118

1sin θ

+ C = − 118csc θ + C = − x

9√4x2 − 9

+ C

18. x = 5 sec θ, dx = 5 sec θ tan θ dθ,

= 375∫sec4 θ dθ

= 125 sec2 θ tan θ + 250∫sec2 θ dθ

= 125 sec2 θ tan θ + 250 tan θ + C

= x2√x2 − 25 + 50

√x2 − 25 + C

19. ex = sin θ, exdx = cos θ dθ,∫cos2 θ dθ =

12

∫(1 + cos 2θ)dθ =

12θ +

14sin 2θ + C =

12sin−1(ex) +

12ex√1− e2x + C

20. u = sin θ,∫

1√2− u2

du = sin−1(sin θ√2

)+ C

21. x = sin θ, dx = cos θ dθ,

5∫ 1

0sin3 θ cos2 θ dθ = 5

[−13cos3 θ +

15cos5 θ

]π/20

= 5(1/3− 1/5) = 2/3

22. x = sin θ, dx = cos θ dθ,∫ π/6

0sec3 θ, dθ=

[12sec θ tan θ +

12ln | sec θ + tan θ|

]π/60

=(122√31√3+12ln(

2√3+1√3

)=13+14ln 3

23. x = sec θ, dx = sec θ tan θ dθ,∫ π/3

π/4

1sec θ

dθ =∫ π/3

π/4cos θ dθ = sin θ

]π/3π/4

= (√3−√2)/2

24. x =√2 sec θ, dx =

√2 sec θ tan θ dθ, 2

∫ π/4

0tan2 θ dθ =

[2 tan θ − 2θ

]π/40

= 2− π/2

January 27, 2005 11:45 L24-CH08 Sheet number 23 Page number 359 black

Exercise Set 8.4 359

25. x =√3 tan θ, dx =

√3 sec2 θ dθ,

19

∫ π/3

π/6

sec θtan4 θ

dθ=19

∫ π/3

π/6

cos3 θsin4 θ

dθ =19

∫ π/3

π/6

1− sin2 θ

sin4 θcos θ dθ =

19

∫ √3/2

1/2

1− u2

u4 du (u = sin θ)

=19

∫ √3/2

1/2(u−4 − u−2)du =

19

[− 13u3 +

1u

]√3/2

1/2=10√3 + 18243

26. x =√3 tan θ, dx =

√3 sec2 θ dθ,

√33

∫ π/3

0

tan3 θ

sec3 θdθ =

√33

∫ π/3

0sin3 θ dθ =

√33

∫ π/3

0

[1− cos2 θ

]sin θ dθ

=√33

[− cos θ + 1

3cos3 θ

]π/30

=√33

[(−12+124

)−(−1 + 1

3

)]= 5√3/72

27. u = x2 + 4, du = 2x dx,

12

∫1udu =

12ln |u|+ C = 1

2ln(x2 + 4) + C; or x = 2 tan θ, dx = 2 sec2 θ dθ,

∫tan θ dθ= ln | sec θ|+ C1 = ln

√x2 + 42

+ C1 = ln(x2 + 4)1/2 − ln 2 + C1

=12ln(x2 + 4) + C with C = C1 − ln 2

28. x = 2 tan θ, dx = 2 sec2 θ dθ,∫2 tan2 θ dθ = 2 tan θ − 2θ + C = x− 2 tan−1 x

2+ C; alternatively∫

x2

x2 + 4dx =

∫dx− 4

∫dx

x2 + 4= x− 2 tan−1 x

2+ C

29. y′ =1x, 1 + (y′)2 = 1 +

1x2 =

x2 + 1x2 ,

L =∫ 2

1

√x2 + 1x2 dx; x = tan θ, dx = sec2 θ dθ,

L =∫ tan−1(2)

π/4

sec3 θtan θ

dθ =∫ tan−1(2)

π/4

tan2 θ + 1tan θ

sec θ dθ =∫ tan−1(2)

π/4(sec θ tan θ + csc θ)dθ

=[sec θ + ln | csc θ − cot θ|

]tan−1(2)

π/4=√5 + ln

(√52− 12

)−[√2 + ln |

√2− 1|

]

=√5−√2 + ln

2 + 2√2

1 +√5

30. y′ = 2x, 1 + (y′)2 = 1 + 4x2,

L =∫ 1

0

√1 + 4x2dx; x =

12tan θ, dx =

12sec2 θ dθ,

L=12

∫ tan−1 2

0sec3 θ dθ =

12

(12sec θ tan θ +

12ln | sec θ + tan θ|

) ]tan−1 2

0

=14(√5)(2) +

14ln |√5 + 2| = 1

2

√5 +

14ln(2 +

√5)

January 27, 2005 11:45 L24-CH08 Sheet number 24 Page number 360 black

360 Chapter 8

31. y′ = 2x, 1 + (y′)2 = 1 + 4x2,

S = 2π∫ 1

0x2√1 + 4x2dx; x =

12tan θ, dx =

12sec2 θ dθ,

S=π

4

∫ tan−1 2

0tan2 θ sec3 θ dθ =

π

4

∫ tan−1 2

0(sec2 θ − 1) sec3 θ dθ = π

4

∫ tan−1 2

0(sec5 θ − sec3 θ)dθ

4

[14sec3 θ tan θ − 1

8sec θ tan θ − 1

8ln | sec θ + tan θ|

]tan−1 2

0=π

32[18√5− ln(2 +

√5)]

32. V = π∫ 1

0y2√1− y2dy; y = sin θ, dy = cos θ dθ,

V = π∫ π/2

0sin2 θ cos2 θ dθ =

π

4

∫ π/2

0sin2 2θ dθ =

π

8

∫ π/2

0(1− cos 4θ)dθ = π

8

(θ − 1

4sin 4θ

)]π/20=π2

16

33.∫

1(x− 2)2 + 1dx = tan

−1 (x− 2) + C 34.∫

1√1− (x− 1)2

dx = sin−1(x− 1) + C

35.∫

1√4− (x− 1)2

dx = sin−1(x− 12

)+ C

36.∫

116(x+ 1/2)2 + 1

dx =14

∫1

(x+ 2)2 + 1dx =

14tan−1(4x+ 2) + C

37.∫

1√(x− 3)2 + 1

dx = ln(x− 3 +

√(x− 3)2 + 1

)+ C

38.∫

x

(x+ 1)2 + 1dx, let u = x+ 1,

∫u− 1u2 + 1

du=∫ (

u

u2 + 1− 1u2 + 1

)du =

12ln(u2 + 1)− tan−1 u+ C

=12ln(x2 + 2x+ 2)− tan−1(x+ 1) + C

39.∫ √

4− (x+ 1)2 dx, let x+ 1 = 2 sin θ,

=∫4 cos2 θ dθ =

∫2(1 + cos 2θ) dθ

= 2θ + sin 2θ + C = 2 sin−1(x+ 12

)+12(x+ 1)

√3− 2x− x2 + C

40.∫

ex√(ex + 1/2)2 + 3/4

dx, let u = ex + 1/2,

∫1√

u2 + 3/4du = sinh−1(2u/

√3) + C = sinh−1

(2ex + 1√

3

)+ C

Alternate solution: let ex + 1/2 =√32tan θ,

∫sec θ dθ= ln | sec θ + tan θ|+ C = ln

(2√e2x + ex + 1√

3+2ex + 1√

3

)+ C1

= ln(2√e2x + ex + 1 + 2ex + 1) + C

January 27, 2005 11:45 L24-CH08 Sheet number 25 Page number 361 black

Exercise Set 8.5 361

41.∫

12(x+ 1)2 + 5

dx =12

∫1

(x+ 1)2 + 5/2dx =

1√10tan−1

√2/5(x+ 1) + C

42.∫

2x+ 34(x+ 1/2)2 + 4

dx, let u = x+ 1/2,

∫2u+ 24u2 + 4

du=12

∫ (u

u2 + 1+

1u2 + 1

)du =

14ln(u2 + 1) +

12tan−1 u+ C

=14ln(x2 + x+ 5/4) +

12tan−1(x+ 1/2) + C

43.∫ 2

1

1√4x− x2

dx =∫ 2

1

1√4− (x− 2)2

dx = sin−1 x− 22

]2

1= π/6

44.∫ 4

0

√4x− x2dx =

∫ 4

0

√4− (x− 2)2dx, let x− 2 = 2 sin θ,

4∫ π/2

−π/2cos2 θ dθ =

[2θ + sin 2θ

]π/2−π/2

= 2π

45. u = sin2 x, du = 2 sinx cosx dx;12

∫ √1− u2 du =

14

[u√1− u2 + sin−1 u

]+ C =

14

[sin2 x

√1− sin4 x+ sin−1(sin2 x)

]+ C

46. u = x sinx, du = (x cosx+ sinx) dx;∫ √1 + u2 du =

12u√1 + u2 +

12sinh−1 u+ C =

12x sinx

√1 + x2 sin2 x+

12sinh−1(x sinx) + C

47. (a) x = 3 sinhu, dx = 3 coshu du,∫du = u+ C = sinh−1(x/3) + C

(b) x = 3 tan θ, dx = 3 sec2 θ dθ,∫sec θ dθ = ln | sec θ + tan θ|+ C = ln

(√x2 + 9/3 + x/3

)+ C

but sinh−1(x/3) = ln(x/3 +

√x2/9 + 1

)= ln

(x/3 +

√x2 + 9/3

)so the results agree.

48. x = coshu, dx = sinhu du,∫sinh2 u du=

12

∫(cosh 2u− 1)du = 1

4sinh 2u− 1

2u+ C

=12sinhu coshu− 1

2u+ C =

12x√x2 − 1− 1

2cosh−1 x+ C

because coshu = x, and sinhu =√cosh2 u− 1 =

√x2 − 1

EXERCISE SET 8.5

1.A

(x− 3) +B

(x+ 4)2.

5x(x− 2)(x+ 2) =

A

x+

B

x− 2 +C

x+ 2

3.2x− 3x2(x− 1) =

A

x+B

x2 +C

x− 1 4.A

x+ 2+

B

(x+ 2)2+

C

(x+ 2)3

5.A

x+B

x2 +C

x3 +Dx+ Ex2 + 2

6.A

x− 1 +Bx+ Cx2 + 6

January 27, 2005 11:45 L24-CH08 Sheet number 26 Page number 362 black

362 Chapter 8

7.Ax+Bx2 + 5

+Cx+D(x2 + 5)2

8.A

x− 2 +Bx+ Cx2 + 1

+Dx+ E(x2 + 1)2

9.1

(x− 4)(x+ 1) =A

x− 4 +B

x+ 1; A =

15, B = −1

5so

15

∫1

x− 4dx−15

∫1

x+ 1dx =

15ln |x− 4| − 1

5ln |x+ 1|+ C = 1

5ln∣∣∣∣x− 4x+ 1

∣∣∣∣+ C

10.1

(x+ 1)(x− 7) =A

x+ 1+

B

x− 7 ; A = −18, B =

18so

−18

∫1

x+ 1dx+

18

∫1

x− 7dx = −18ln |x+ 1|+ 1

8ln |x− 7|+ C = 1

8ln∣∣∣∣x− 7x+ 1

∣∣∣∣+ C

11.11x+ 17

(2x− 1)(x+ 4) =A

2x− 1 +B

x+ 4; A = 5, B = 3 so

5∫

12x− 1dx+ 3

∫1

x+ 4dx =

52ln |2x− 1|+ 3 ln |x+ 4|+ C

12.5x− 5

(x− 3)(3x+ 1) =A

x− 3 +B

3x+ 1; A = 1, B = 2 so∫

1x− 3dx+ 2

∫1

3x+ 1dx = ln |x− 3|+ 2

3ln |3x+ 1|+ C

13.2x2 − 9x− 9x(x+ 3)(x− 3) =

A

x+

B

x+ 3+

C

x− 3 ; A = 1, B = 2, C = −1 so∫1xdx+ 2

∫1

x+ 3dx−

∫1

x− 3 dx = ln |x|+ 2 ln |x+ 3| − ln |x− 3|+ C = ln∣∣∣∣x(x+ 3)2x− 3

∣∣∣∣+ CNote that the symbol C has been recycled; to save space this recycling is usually not mentioned.

14.1

x(x+ 1)(x− 1) =A

x+

B

x+ 1+

C

x− 1 ; A = −1, B =12, C =

12so

−∫1xdx+

12

∫1

x+ 1dx+

12

∫1

x− 1 dx= − ln |x|+12ln |x+ 1|+ 1

2ln |x− 1|+ C

=12ln∣∣∣∣ (x+ 1)(x− 1)x2

∣∣∣∣+ C = 12ln|x2 − 1|x2 + C

15.x2 − 8x+ 3

= x− 3 + 1x+ 3

,∫ (

x− 3 + 1x+ 3

)dx =

12x2 − 3x+ ln |x+ 3|+ C

16.x2 + 1x− 1 = x+ 1 +

2x− 1 ,

∫ (x+ 1 +

2x− 1

)dx =

12x2 + x+ 2 ln |x− 1|+ C

17.3x2 − 10x2 − 4x+ 4 = 3 +

12x− 22x2 − 4x+ 4,

12x− 22(x− 2)2 =

A

x− 2 +B

(x− 2)2 ; A = 12, B = 2 so∫3dx+ 12

∫1

x− 2dx+ 2∫

1(x− 2)2 dx = 3x+ 12 ln |x− 2| − 2/(x− 2) + C

18.x2

x2 − 3x+ 2 = 1 +3x− 2

x2 − 3x+ 2,3x− 2

(x− 1)(x− 2) =A

x− 1 +B

x− 2 ; A = −1, B = 4 so∫dx−

∫1

x− 1dx+ 4∫

1x− 2dx = x− ln |x− 1|+ 4 ln |x− 2|+ C

January 27, 2005 11:45 L24-CH08 Sheet number 27 Page number 363 black

Exercise Set 8.5 363

19.x5 + x2 + 2x3 − x = x2 + 1 +

x2 + x+ 2x3 − x ,

x2 + x+ 2x(x+ 1)(x− 1) =

A

x+

B

x+ 1+

C

x− 1 ; A = −2, B = 1, C = 2 so∫(x2 + 1)dx−

∫2xdx+

∫1

x+ 1dx+

∫2

x− 1dx

=13x3 + x− 2 ln |x|+ ln |x+ 1|+ 2 ln |x− 1|+ C = 1

3x3 + x+ ln

∣∣∣∣ (x+ 1)(x− 1)2x2

∣∣∣∣+ C20.

x5 − 4x3 + 1x3 − 4x = x2 +

1x3 − 4x ,

1x(x+ 2)(x− 2) =

A

x+

B

x+ 2+

C

x− 2 ; A = −14, B = +

18, C = +

18so

∫x2 dx− 1

4

∫1xdx+

18

∫1

x+ 2dx+

18

∫1

x− 2 dx

=13x3 − 1

4ln |x|+ 1

8ln |x+ 2|+ 1

8ln |x− 2|+ C

21.2x2 + 3x(x− 1)2 =

A

x+

B

x− 1 +C

(x− 1)2 ; A = 3, B = −1, C = 5 so

3∫1xdx−

∫1

x− 1 dx+ 5∫

1(x− 1)2 dx = 3 ln |x| − ln |x− 1| − 5/(x− 1) + C

22.3x2 − x+ 1x2(x− 1) =

A

x+B

x2 +C

x− 1 ; A = 0, B = −1, C = 3 so

−∫1x2 dx+ 3

∫1

x− 1dx = 1/x+ 3 ln |x− 1|+ C

23.2x2 − 10x+ 4(x+ 1)(x− 3)2 =

A

x+ 1+

B

x− 3 +C

(x− 3)2 ; A = 1, B = 1, C = −2 so∫1

x+ 1dx+

∫1

x− 3 dx−∫

2(x− 3)2 dx = ln |x+ 1|+ ln |x− 3|+

2x− 3 + C1

24.2x2 − 2x− 1x2(x− 1) =

A

x+B

x2 +C

x− 1 ; A = 3, B = 1, C = −1 so

3∫1xdx+

∫1x2 dx−

∫1

x− 1 dx = 3 ln |x| −1x− ln |x− 1|+ C

25.x2

(x+ 1)3=

A

x+ 1+

B

(x+ 1)2+

C

(x+ 1)3; A = 1, B = −2, C = 1 so

∫1

x+ 1dx−

∫2

(x+ 1)2dx+

∫1

(x+ 1)3dx = ln |x+ 1|+ 2

x+ 1− 12(x+ 1)2

+ C

26.2x2 + 3x+ 3(x+ 1)3

=A

x+ 1+

B

(x+ 1)2+

C

(x+ 1)3; A = 2, B = −1, C = 2 so

2∫

1x+ 1

dx−∫

1(x+ 1)2

dx+ 2∫

1(x+ 1)3

dx = 2 ln |x+ 1|+ 1x+ 1

− 1(x+ 1)2

+ C

January 27, 2005 11:45 L24-CH08 Sheet number 28 Page number 364 black

364 Chapter 8

27.2x2 − 1

(4x− 1)(x2 + 1)=

A

4x− 1 +Bx+ Cx2 + 1

; A = −14/17, B = 12/17, C = 3/17 so∫

2x2 − 1(4x− 1)(x2 + 1)

dx = − 734ln |4x− 1|+ 6

17ln(x2 + 1) +

317tan−1 x+ C

28.1

x(x2 + 2)=A

x+Bx+ Cx2 + 2

;A =12, B = −1

2, C = 0 so

∫1

x3 + 2xdx =

12ln |x| − 1

4ln(x2 + 2) + C =

14ln

x2

x2 + 2+ C

29.x3 + 3x2 + x+ 9(x2 + 1)(x2 + 3)

=Ax+Bx2 + 1

+Cx+Dx2 + 3

; A = 0, B = 3, C = 1, D = 0 so

∫x3 + 3x2 + x+ 9(x2 + 1)(x2 + 3)

dx = 3 tan−1 x+12ln(x2 + 3) + C

30.x3 + x2 + x+ 2(x2 + 1)(x2 + 2)

=Ax+Bx2 + 1

+Cx+Dx2 + 2

; A = D = 0, B = C = 1 so

∫x3 + x2 + x+ 2(x2 + 1)(x2 + 2)

dx = tan−1 x+12ln(x2 + 2) + C

31.x3 − 2x2 + 2x− 2

x2 + 1= x− 2 + x

x2 + 1,

∫x3 − 3x2 + 2x− 3

x2 + 1dx =

12x2 − 2x+ 1

2ln(x2 + 1) + C

32.x4 + 6x3 + 10x2 + x

x2 + 6x+ 10= x2 +

x

x2 + 6x+ 10,∫

x

x2 + 6x+ 10dx=

∫x

(x+ 3)2 + 1dx =

∫u− 3u2 + 1

du, u = x+ 3

=12ln(u2 + 1)− 3 tan−1 u+ C1

so∫x4 + 6x3 + 10x2 + x

x2 + 6x+ 10dx =

13x3 +

12ln(x2 + 6x+ 10)− 3 tan−1(x+ 3) + C

33. Let x = sin θ to get∫

1x2 + 4x− 5 dx, and

1(x+ 5)(x− 1) =

A

x+ 5+

B

x− 1 ; A = −1/6,

B = 1/6 so we get −16

∫1

x+ 5dx+

16

∫1

x− 1 dx =16ln∣∣∣∣x− 1x+ 5

∣∣∣∣+ C = 16ln(1− sin θ5 + sin θ

)+ C.

34. Let x = et; then∫

et

e2t − 4dt =∫

1x2 − 4 dx,

1(x+ 2)(x− 2) =

A

x+ 2+

B

x− 2 ; A = −1/4, B = 1/4 so

−14

∫1

x+ 2dx+

14

∫1

x− 2 dx =14ln∣∣∣∣x− 2x+ 2

∣∣∣∣+ C = 14ln∣∣∣∣et − 2et + 2

∣∣∣∣+ C.

January 27, 2005 11:45 L24-CH08 Sheet number 29 Page number 365 black

Exercise Set 8.5 365

35. V = π∫ 2

0

x4

(9− x2)2dx,

x4

x4 − 18x2 + 81= 1 +

18x2 − 81x4 − 18x2 + 81

,

18x2 − 81(9− x2)2

=18x2 − 81

(x+ 3)2(x− 3)2 =A

x+ 3+

B

(x+ 3)2+

C

x− 3 +D

(x− 3)2 ;

A = −94, B =

94, C =

94, D =

94so

V = π[x− 9

4ln |x+ 3| − 9/4

x+ 3+94ln |x− 3| − 9/4

x− 3

]2

0= π

(195− 94ln 5)

36. Let u = ex to get∫ ln 5

− ln 5

dx

1 + ex=∫ ln 5

− ln 5

exdx

ex(1 + ex)=∫ 5

1/5

du

u(1 + u),

1u(1 + u)

=A

u+

B

1 + u; A = 1, B = −1;

∫ 5

1/5

du

u(1 + u)= (lnu− ln(1 + u))

]5

1/5= ln 5

37.x2 + 1

(x2 + 2x+ 3)2=

Ax+Bx2 + 2x+ 3

+Cx+D

(x2 + 2x+ 3)2; A = 0, B = 1, C = D = −2 so

∫x2 + 1

(x2 + 2x+ 3)2dx=

∫1

(x+ 1)2 + 2dx−

∫2x+ 2

(x2 + 2x+ 3)2dx

=1√2tan−1 x+ 1√

2+ 1/(x2 + 2x+ 3) + C

38.x5 + x4 + 4x3 + 4x2 + 4x+ 4

(x2 + 2)3=Ax+Bx2 + 2

+Cx+D(x2 + 2)2

+Ex+ F(x2 + 2)3

;

A = B = 1, C = D = E = F = 0 so∫x+ 1x2 + 2

dx =12ln(x2 + 2) +

1√2tan−1(x/

√2) + C

39. x4 − 3x3 − 7x2 + 27x− 18 = (x− 1)(x− 2)(x− 3)(x+ 3),1

(x− 1)(x− 2)(x− 3)(x+ 3) =A

x− 1 +B

x− 2 +C

x− 3 +D

x+ 3;

A = 1/8, B = −1/5, C = 1/12, D = −1/120 so∫dx

x4 − 3x3 − 7x2 + 27x− 18 =18ln |x− 1| − 1

5ln |x− 2|+ 1

12ln |x− 3| − 1

120ln |x+ 3|+ C

40. 16x3 − 4x2 + 4x− 1 = (4x− 1)(4x2 + 1),1

(4x− 1)(4x2 + 1)=

A

4x− 1 +Bx+ C4x2 + 1

; A = 4/5, B = −4/5, C = −1/5 so∫

dx

16x3 − 4x2 + 4x− 1 =15ln |4x− 1| − 1

10ln(4x2 + 1)− 1

10tan−1(2x) + C

41. Let u = x2, du = 2x dx,∫ 1

0

x

x4 + 1dx =

12

∫ 1

0

11 + u2 du =

12tan−1 u

]1

0=12π

4=π

8.

42.1

a2 − x2 =A

a− x +B

a+ x;A =

12a,B =

12aso

12a

∫ (1

a− x +1

a+ x

)dx =

12a(− ln |a− x|+ ln |a+ x| ) + C = 1

2aln∣∣∣∣a+ xa− x

∣∣∣∣+ C

January 27, 2005 11:45 L24-CH08 Sheet number 30 Page number 366 black

366 Chapter 8

43. If the polynomial has distinct roots r1, r2, r1 �= r2, then the partial fraction decomposition will

contain terms of the formA

x− r1,B

x− r2, and they will give logarithms and no inverse tangents. If

there are two roots not distinct, say x = r, then the termsA

x− r ,B

(x− r)2 will appear, and neither

will give an inverse tangent term. The only other possibility is no real roots, and the integrand

can be written in the form1

a(x+ b

2a

)2+ c− b2

4a

, which will yield an inverse tangent, specifically

of the form tan−1[A

(x+

b

2a

)]for some constant A.

44. Since there are no inverse tangent terms, the roots are real. Since there are no logarithmic terms,

there are no terms of the form1

x− r , so the only terms that can arise in the partial fraction

decomposition form is one like1

45. Yes, for instance the integrand1

x2 + 1, whose integral is precisely tan−1 x+ C.

EXERCISE SET 8.6

1. Formula (60):49

[3x+ ln |−1 + 3x|

]+ C 2. Formula (62):

125

[4

4− 5x + ln |4− 5x|]+ C

3. Formula (65):15ln∣∣∣∣ x

5 + 2x

∣∣∣∣+ C 4. Formula (66): − 1x− 5 ln

∣∣∣∣1− 5xx∣∣∣∣+ C

5. Formula (102):15(x− 1)(2x+ 3)3/2 + C 6. Formula (105):

23(−x− 4)

√2− x+ C

7. Formula (108):12ln∣∣∣∣√4− 3x− 2√4− 3x+ 2

∣∣∣∣+ C 8. Formula (108): tan−1

√3x− 42

+ C

9. Formula (69):18ln∣∣∣∣x+ 4x− 4

∣∣∣∣+ C 10. Formula (70):16ln∣∣∣∣x− 3x+ 3

∣∣∣∣+ C11. Formula (73):

x

2

√x2 − 3− 3

2ln∣∣∣x+√x2 − 3

∣∣∣+ C12. Formula (94): −

√x2 − 5x

+ ln(x+√x2 − 5) + C

13. Formula (95):x

2

√x2 + 4− 2 ln(x+

√x2 + 4) + C

14. Formula (90): −√x2 − 22x

+ C 15. Formula (74):x

2

√9− x2 +

92sin−1 x

3+ C

16. Formula (80): −√4− x2

x− sin−1 x

2+ C

January 27, 2005 11:45 L24-CH08 Sheet number 31 Page number 367 black

Exercise Set 8.6 367

17. Formula (79):√4− x2 − 2 ln

∣∣∣∣∣2 +√4− x2

x

∣∣∣∣∣+ C

18. Formula (117): −√6x− x2

3x+ C 19. Formula (38): − 1

14sin(7x) +

12sinx+ C

20. Formula (40): − 114cos(7x) +

16cos(3x) + C

21. Formula (50):x4

16[4 lnx− 1] + C 22. Formula (50): −2 lnx+ 2√

x

23. Formula (42):e−2x

13(−2 sin(3x)− 3 cos(3x)) + C

24. Formula (43):ex

5(cos(2x) + 2 sin(2x)) + C

25. u = e2x, du = 2e2xdx, Formula (62):12

∫u du

(4− 3u)2 =118

[4

4− 3e2x + ln∣∣4− 3e2x∣∣]+ C

26. u = cos 2x, du = −2 sin 2xdx, Formula (65): −∫

du

2u(3− u) = −16ln∣∣∣∣ cos 2x3− cos 2x

∣∣∣∣+ C27. u = 3

√x, du =

32√xdx, Formula (68):

23

∫du

u2 + 4=13tan−1 3

√x

2+ C

28. u = sin 4x, du = 4 cos 4xdx, Formula (68):14

∫du

9 + u2 =112tan−1 sin 4x

3+ C

29. u = 2x, du = 2dx, Formula (76):12

∫du√u2 − 9

=12ln∣∣∣2x+√4x2 − 9

∣∣∣+ C30. u =

√2x2, du = 2

√2xdx, Formula (72):

12√2

∫ √u2 + 3 du =

x2

4

√2x4 + 3 +

34√2ln(√2x2 +

√2x4 + 3

)+ C

31. u = 2x2, du = 4xdx, u2du = 16x5 dx, Formula (81):

14

∫u2 du√2− u2

= −x2

4

√2− 4x4 +

14sin−1(

√2x2) + C

32. u = 2x, du = 2dx, Formula (83): 2∫

du

u2√3− u2

= − 13x

√3− 4x2 + C

33. u = lnx, du = dx/x, Formula (26):∫sin2 u du =

12lnx+

14sin(2 lnx) + C

34. u = e−2x, du = −2e−2x, Formula (27): −12

∫cos2 u du = −1

4e−2x − 1

8sin(2e−2x)+ C

35. u = −2x, du = −2dx, Formula (51): 14

∫ueu du =

14(−2x− 1)e−2x + C

January 27, 2005 11:45 L24-CH08 Sheet number 32 Page number 368 black

368 Chapter 8

36. u = 3x+1, du = 3 dx, Formula (11):13

∫lnu du =

13(u lnu−u)+C = 1

3(3x+1)[ln(3x+1)−1]+C

37. u = sin 3x, du = 3 cos 3x dx, Formula (67):13

∫du

u(u+ 1)2=13

[1

1 + sin 3x+ ln

∣∣∣∣ sin 3x1 + sin 3x

∣∣∣∣]+C

38. u = lnx, du =1xdx, Formula (105):

∫u du√4u− 1

=112(2 lnx+ 1)

√4 lnx− 1 + C

39. u = 4x2, du = 8xdx, Formula (70):18

∫du

u2 − 1 =116ln∣∣∣∣4x2 − 14x2 + 1

∣∣∣∣+ C

40. u = 2ex, du = 2exdx, Formula (69):12

∫du

3− u2 =14√3ln

∣∣∣∣∣2ex +√3

2ex −√3

∣∣∣∣∣+ C41. u = 2ex, du = 2exdx, Formula (74):

12

∫ √3− u2 du =

14u√3− u2 +

34sin−1(u/

√3) + C =

12ex√3− 4e2x + 3

4sin−1(2ex/

√3) + C

42. u = 3x, du = 3dx, Formula (80):

3∫ √

4− u2du

u2 = −3√4− u2

u− 3 sin−1(u/2) + C = −

√4− 9x2

x− 3 sin−1(3x/2) + C

43. u = 3x, du = 3dx, Formula (112):

13

∫ √53u− u2 du =

16

(u− 5

6

)√53u− u2 +

25216

sin−1(u− 55

)+ C

=18x− 536

√5x− 9x2 +

25216

sin−1(18x− 55

)+ C

44. u =√5x, du =

√5 dx, Formula (117):

∫du

u√(u/√5)− u2

= −

√(u/√5)− u2

u/(2√5)

+ C = −2√x− 5x2

x+ C

45. u = 2x, du = 2dx, Formula (44):∫u sinu du = (sinu− u cosu) + C = sin 2x− 2x cos 2x+ C

46. u =√x, u2 = x, 2udu = dx, Formula (45): 2

∫u cosu du = 2 cos

√x+ 2

√x sin

√x+ C

47. u = −√x, u2 = x, 2udu = dx, Formula (51): 2

∫ueudu = −2(

√x+ 1)e−

√x + C

48. u = 2 + x2, du = 2x dx, Formula (11):

12

∫lnu du =

12(u lnu− u) + C = 1

2(2 + x2) ln(2 + x2)− 1

2(2 + x2) + C

49. x2 + 6x− 7 = (x+ 3)2 − 16;u = x+ 3, du = dx, Formula (70):∫du

u2 − 16 =18ln∣∣∣∣u− 4u+ 4

∣∣∣∣+ C = 18ln∣∣∣∣x− 1x+ 7

∣∣∣∣+ C

January 27, 2005 11:45 L24-CH08 Sheet number 33 Page number 369 black

Exercise Set 8.6 369

50. x2 + 2x− 3 = (x+ 1)2 − 4, u = x+ 1, du = dx, Formula (77):∫ √4− u2 du =

12u√4− u2 + 2 sin−1(u/2) + C

=12(x+ 1)

√3− 2x− x2 + 2 sin−1((x+ 1)/2) + C

51. x2 − 4x− 5 = (x− 2)2 − 9, u = x− 2, du = dx, Formula (77):∫u+ 2√9− u2

du =∫

u du√9− u2

+ 2∫

du√9− u2

= −√9− u2 + 2 sin−1 u

3+ C

= −√5 + 4x− x2 + 2 sin−1

(x− 23

)+ C

52. x2 + 6x+ 13 = (x+ 3)2 + 4, u = x+ 3, du = dx, Formula (71):∫(u− 3) duu2 + 4

=12ln(u2 + 4)− 3

2tan−1(u/2) + C =

12ln(x2 + 6x+ 13)− 3

2tan−1((x+ 3)/2) + C

53. u =√x− 2, x = u2 + 2, dx = 2u du;∫

2u2(u2 + 2)du = 2∫(u4 + 2u2)du =

25u5 +

43u3 + C =

25(x− 2)5/2 + 4

3(x− 2)3/2 + C

54. u =√x+ 1, x = u2 − 1, dx = 2u du;

2∫(u2 − 1)du = 2

3u3 − 2u+ C = 2

3(x+ 1)3/2 − 2

√x+ 1 + C

55. u =√x3 + 1, x3 = u2 − 1, 3x2dx = 2u du;

23

∫u2(u2 − 1)du = 2

3

∫(u4 − u2)du =

215u5 − 2

9u3 + C =

215(x3 + 1)5/2 − 2

9(x3 + 1)3/2 + C

56. u =√x3 − 1, x3 = u2 + 1, 3x2dx = 2u du;

23

∫1

u2 + 1du =

23tan−1 u+ C =

23tan−1

√x3 − 1 + C

57. u = x1/3, x = u3, dx = 3u2 du;∫3u2

u3 − udu= 3∫

u

u2 − 1du = 3∫ [

12(u+ 1)

+1

2(u− 1)

]du

=32ln |x1/3 + 1|+ 3

2ln |x1/3 − 1|+ C

58. u = x1/6, x = u6, dx = 6u5du;∫6u5

u3 + u2 du= 6∫

u3

u+ 1du = 6

∫ [u2 − u+ 1− 1

u+ 1

]du

= 2x1/2 − 3x1/3 + 6x1/6 − 6 ln(x1/6 + 1) + C

59. u = x1/4, x = u4, dx = 4u3du; 4∫

1u(1− u)du = 4

∫ [1u+

11− u

]du = 4 ln

x1/4

|1− x1/4| + C

60. u = x1/2, x = u2, dx = 2u du;∫2u2

u2 + 1du = 2

∫u2 + 1− 1u2 + 1

du = 2u− 2 tan−1 u+ C = 2√x− 2 tan−1√x+ C

January 27, 2005 11:45 L24-CH08 Sheet number 34 Page number 370 black

370 Chapter 8

61. u = x1/6, x = u6, dx = 6u5du;

6∫

u3

u− 1du = 6∫ [

u2 + u+ 1 +1

u− 1

]du = 2x1/2 + 3x1/3 + 6x1/6 + 6 ln |x1/6 − 1|+ C

62. u =√x, x = u2, dx = 2u du;

−2∫u2 + uu− 1 du = −2

∫ (u+ 2 +

2u− 1

)du = −x− 4

√x− 4 ln |

√x− 1|+ C

63. u =√1 + x2, x2 = u2 − 1, 2x dx = 2u du, x dx = u du;∫

(u2 − 1)du = 13(1 + x2)3/2 − (1 + x2)1/2 + C

64. u = (x+ 3)1/5, x = u5 − 3, dx = 5u4du;

5∫(u8 − 3u3)du =

59(x+ 3)9/5 − 15

4(x+ 3)4/5 + C

65.∫

1

1 +2u

1 + u2 +1− u2

1 + u2

21 + u2 du =

∫1

u+ 1du = ln | tan(x/2) + 1|+ C

66.∫

1

2 +2u

1 + u2

21 + u2 du=

∫1

u2 + u+ 1du

=∫

1(u+ 1/2)2 + 3/4

du =2√3tan−1

(2 tan(x/2) + 1√

3

)+ C

67. u = tan(θ/2),∫

1− cos θ =∫1u2 du = −

1u+ C = − cot(θ/2) + C

68. u = tan(x/2),∫2

3u2 + 8u− 3du=23

∫1

(u+ 4/3)2 − 25/9du =23

∫1

z2 − 25/9dz (z = u+ 4/3)

=15ln∣∣∣∣z − 5/3z + 5/3

∣∣∣∣+ C = 15ln∣∣∣∣ tan(x/2)− 1/3tan(x/2) + 3

∣∣∣∣+ C69. u = tan(x/2),

12

∫1− u2

udu =

12

∫(1/u− u)du = 1

2ln | tan(x/2)| − 1

4tan2(x/2) + C

70. u = tan(x/2),∫1− u2

1 + u2 du = −u+ 2 tan−1 u+ C = x− tan(x/2) + C

71.∫ x

2

1t(4− t) dt=

14ln

t

4− t

]x2(Formula (65), a = 4, b = −1)

=14

[ln

x

4− x − ln 1]=14ln

x

4− x,14ln

x

4− x = 0.5, lnx

4− x = 2,

x

4− x = e2, x = 4e2 − e2x, x(1 + e2) = 4e2, x = 4e2/(1 + e2) ≈ 3.523188312

January 27, 2005 11:45 L24-CH08 Sheet number 35 Page number 371 black

Exercise Set 8.6 371

72.∫ x

1

1t√2t− 1

dt= 2 tan−1√2t− 1]x

1(Formula (108), a = −1, b = 2)

= 2(tan−1√2x− 1− tan−1 1

)= 2

(tan−1√2x− 1− π/4

),

2(tan−1√2x− 1− π/4) = 1, tan−1√2x− 1 = 1/2 + π/4,√2x− 1 = tan(1/2 + π/4),

x = [1 + tan2(1/2 + π/4)]/2 ≈ 6.307993516

73. A =∫ 4

0

√25− x2 dx=

(12x√25− x2 +

252sin−1 x

5

)]4

0(Formula (74), a = 5)

= 6 +252sin−1 4

5≈ 17.59119023

74. A =∫ 2

2/3

√9x2 − 4 dx; u = 3x,

A=13

∫ 6

2

√u2 − 4 du = 1

3

(12u√u2 − 4− 2 ln

∣∣∣u+√u2 − 4∣∣∣)]6

2(Formula (73), a2 = 4)

=13

(3√32− 2 ln(6 +

√32) + 2 ln 2

)= 4√2− 2

3ln(3 + 2

√2) ≈ 4.481689467

75. A =∫ 1

0

125− 16x2 dx; u = 4x,

A =14

∫ 4

0

125− u2 du =

140ln∣∣∣∣u+ 5u− 5

∣∣∣∣]4

0=140ln 9 ≈ 0.054930614 (Formula (69), a = 5)

76. A =∫ 4

1

√x lnx dx=

49x3/2

(32lnx− 1

)]4

1(Formula (50), n = 1/2)

=49(12 ln 4− 7) ≈ 4.282458815

77. V = 2π∫ π/2

0x cosx dx = 2π(cosx+ x sinx)

]π/20

= π(π − 2) ≈ 3.586419094 (Formula (45))

78. V = 2π∫ 8

4x√x− 4 dx= 4π

15(3x+ 8)(x− 4)3/2

]8

4(Formula (102), a = −4, b = 1)

=102415

π ≈ 214.4660585

79. V = 2π∫ 3

0xe−xdx; u = −x,

V = 2π∫ −3

0ueudu = 2πeu(u− 1)

]−3

0= 2π(1− 4e−3) ≈ 5.031899801 (Formula (51))

80. V = 2π∫ 5

1x lnx dx =

π

2x2(2 lnx− 1)

]5

1

= π(25 ln 5− 12) ≈ 88.70584621 (Formula (50), n = 1)

January 27, 2005 11:45 L24-CH08 Sheet number 36 Page number 372 black

372 Chapter 8

81. L =∫ 2

0

√1 + 16x2 dx; u = 4x,

L =14

∫ 8

0

√1 + u2 du=

14

(u

2

√1 + u2 +

12ln(u+

√1 + u2

))]8

0(Formula (72), a2 = 1)

=√65 +

18ln(8 +

√65) ≈ 8.409316783

82. L =∫ 3

1

√1 + 9/x2 dx=

∫ 3

1

√x2 + 9x

dx =

(√x2 + 9− 3 ln

∣∣∣∣∣3 +√x2 + 9x

∣∣∣∣∣)]3

1

= 3√2−√10 + 3 ln

3 +√10

1 +√2≈ 3.891581644 (Formula (89), a = 3)

83. S = 2π∫ π

0(sinx)

√1 + cos2 x dx; u = cosx,

S = −2π∫ −1

1

√1 + u2 du= 4π

∫ 1

0

√1 + u2 du = 4π

(u

2

√1 + u2 +

12ln(u+

√1 + u2

))]1

0a2 = 1

= 2π[√2 + ln(1 +

√2)]≈ 14.42359945 (Formula (72))

84. S = 2π∫ 4

1

1x

√1 + 1/x4 dx = 2π

∫ 4

1

√x4 + 1x3 dx; u = x2,

S = π∫ 16

1

√u2 + 1u2 du= π

(−√u2 + 1u

+ ln(u+

√u2 + 1

))]16

1

= π

(√2−√25716

+ ln16 +

√257

1 +√2

)≈ 9.417237485 (Formula (93), a2 = 1)

85. (a) s(t) = 2 +∫ t

020 cos6 u sin3 u du

= −209sin2 t cos7 t− 40

63cos7 t+

16663

(b)

3 6 9 12 15

1

2

3

4

t

s(t)

86. (a) v(t) =∫ t

0a(u) du = − 1

10e−t cos 2t+

15e−t sin 2t+

174e−t cos 6t− 3

37e−t sin 6t+

110− 174

s(t) = 10 +∫ t

0v(u) du

= − 350e−t cos 2t− 2

25e−t sin 2t+

352738

e−t cos 6t+61369

e−t sin 6t+16185

t+34386634225

January 27, 2005 11:45 L24-CH08 Sheet number 37 Page number 373 black

Exercise Set 8.6 373

(b)

3 6 9 12 15

1

2

3

4

t

s(t)

87. (a)∫secx dx =

∫1

cosxdx =

∫2

1− u2 du = ln∣∣∣∣1 + u1− u

∣∣∣∣+ C = ln∣∣∣∣1 + tan(x/2)1− tan(x/2)

∣∣∣∣+ C= ln

{∣∣∣∣cos(x/2) + sin(x/2)cos(x/2)− sin(x/2)

∣∣∣∣∣∣∣∣cos(x/2) + sin(x/2)cos(x/2) + sin(x/2)

∣∣∣∣}+ C = ln

∣∣∣∣1 + sinxcosx

∣∣∣∣+ C= ln |secx+ tanx|+ C

(b) tan(π4+x

2

)=tan

π

4+ tan

x

21− tan π

4tan

x

2

=1 + tan

x

21− tan x

2

88.∫cscx dx =

∫1sinx

dx =∫1/u du = ln | tan(x/2)|+ C but

ln | tan(x/2)| = 12lnsin2(x/2)cos2(x/2)

=12ln(1− cosx)/2(1 + cosx)/2

=12ln1− cosx1 + cosx

; also,

1− cosx1 + cosx

=1− cos2 x(1 + cosx)2

=1

(cscx+ cotx)2so12ln1− cosx1 + cosx

= − ln | cscx+ cotx|

89. Let u = tanh(x/2) then cosh(x/2) = 1/ sech(x/2) = 1/√1− tanh2(x/2) = 1/

√1− u2,

sinh(x/2) = tanh(x/2) cosh(x/2) = u/√1− u2, so sinhx = 2 sinh(x/2) cosh(x/2) = 2u/(1− u2),

coshx = cosh2(x/2) + sinh2(x/2) = (1 + u2)/(1− u2), x = 2 tanh−1 u, dx = [2/(1− u2)]du;∫dx

2 coshx+ sinhx=∫

1u2 + u+ 1

du =2√3tan−1 2u+ 1√

3+ C =

2√3tan−1 2 tanh(x/2) + 1√

3+ C.

90. Let u = x4 to get14

∫1√1− u2

du =14sin−1 u+ C =

14sin−1(x4) + C.

91.∫(cos32 x sin30 x− cos30 x sin32 x)dx=

∫cos30 x sin30 x(cos2 x− sin2 x)dx

=1230

∫sin30 2x cos 2x dx =

sin31 2x31(231)

+ C

92.∫ √

x−√x2 − 4dx = 1√

2

∫(√x+ 2−

√x− 2)dx =

√23[(x+ 2)3/2 − (x− 2)3/2] + C

93.∫

1x10(1 + x−9)

dx = −19

∫1udu = −1

9ln |u|+ C = −1

9ln |1 + x−9|+ C

January 27, 2005 11:45 L24-CH08 Sheet number 38 Page number 374 black

374 Chapter 8

94. (a) (x+ 4)(x− 5)(x2 + 1)2;A

x+ 4+

B

x− 5 +Cx+Dx2 + 1

+Ex+ F(x2 + 1)2

(b) − 3x+ 4

+2

x− 5 −x− 2x2 + 1

− 3(x2 + 1)2

(c) −3 ln |x+ 4|+ 2 ln |x− 5|+ 2 tan−1 x− 12ln(x2 + 1)− 3

2

(x

x2 + 1+ tan−1 x

)+ C

EXERCISE SET 8.7

1. exact value = 14/3 ≈ 4.666666667(a) 4.667600663, |EM | ≈ 0.000933996(b) 4.664795679, |ET | ≈ 0.001870988(c) 4.666651630, |ES | ≈ 0.000015037

2. exact value = 2

(a) 1.998377048, |EM | ≈ 0.001622952(b) 2.003260982, |ET | ≈ 0.003260982(c) 2.000072698, |ES | ≈ 0.000072698

3. exact value = 2

(a) 2.008248408, |EM | ≈ 0.008248408(b) 1.983523538, |ET | ≈ 0.016476462(c) 2.000109517, |ES | ≈ 0.000109517

4. exact value = sin(1) ≈ 0.841470985(a) 0.841821700, |EM | ≈ 0.000350715(b) 0.840769642, |ET | ≈ 0.000701343(c) 0.841471453, |ES | ≈ 0.000000468

5. exact value = e−1 − e−4 ≈ 0.3495638023(a) 0.3482563710, |EM | ≈ 0.0013074313(b) 0.3521816066, |ET | ≈ 0.0026178043(c) 0.3495793657, |ES | ≈ 0.0000155634

6. exact value =12ln 5 ≈ 0.804718956

(a) 0.801605339, |EM | ≈ 0.003113617(b) 0.811019505, |ET | ≈ 0.006300549(c) 0.805041497, |ES | ≈ 0.000322541

7. f(x) =√x− 1, f ′′(x) = −1

4(x− 1)−3/2, f (4)(x) = −15

16(x− 1)−7/2; K2 = 1/4, K4 = 15/16

(a) |EM | ≤272400

(1/4) = 0.002812500 (b) |ET | ≤271200

(1/4) = 0.005625000

(c) |ES | ≤243

180× 104 (15/16) ≈ 0.000126563

8. f(x) = 1/√x, f ′′(x) =

34x−5/2, f (4)(x) =

10516x−9/2; K2 = 3/4, K4 = 105/16

(a) |EM | ≤272400

(3/4) = 0.008437500 (b) |ET | ≤271200

(3/4) = 0.016875000

(c) |ES | ≤243

180× 104 (105/16) ≈ 0.000885938

9. f(x) = sinx, f ′′(x) = − sinx, f (4)(x) = sinx; K2 = K4 = 1

(a) |EM | ≤π3

2400(1) ≈ 0.012919282 (b) |ET | ≤

π3

1200(1) ≈ 0.025838564

(c) |ES | ≤π5

180× 104 (1) ≈ 0.000170011

10. f(x) = cosx, f ′′(x) = − cosx, f (4)(x) = cosx; K2 = K4 = 1

(a) |EM | ≤12400

(1) ≈ 0.000416667 (b) |ET | ≤11200

(1) ≈ 0.000833333

(c) |ES | ≤1

180× 104 (1) ≈ 0.000000556

January 27, 2005 11:45 L24-CH08 Sheet number 39 Page number 375 black

Exercise Set 8.7 375

11. f(x) = e−x, f ′′(x) = f (4)(x) = e−x; K2 = K4 = e−1

(a) |EM | ≤9800

(e−1) ≈ 0.001226265 (b) |ET | ≤9400

(e−1) ≈ 0.002452530

(c) |ES | ≤27

2× 105 (e−1) ≈ 0.000049664

12. f(x) = 1/(2x+ 1), f ′′(x) = 8(2x+ 1)−3, f (4)(x) = 384(2x+ 1)−5; K2 = 8, K4 = 384

(a) |EM | ≤82400

(8) ≈ 0.026666667 (b) |ET | ≤81200

(8) ≈ 0.053333333

(c) |ES | ≤32

180× 104 (384) ≈ 0.006826667

13. (a) n >

[(27)(1/4)

(24)(5× 10−4)

]1/2

≈ 23.7; n = 24 (b) n >

[(27)(1/4)

(12)(5× 10−4)

]1/2≈ 33.5; n=34

(c) n >

[(243)(15/16)(180)(5× 10−4)

]1/4

≈ 7.1; n = 8

14. (a) n >

[(27)(3/4)

(24)(5× 10−4)

]1/2

≈ 41.1; n = 42 (b) n >

[(27)(3/4)

(12)(5× 10−4)

]1/2≈ 58.1; n=59

(c) n >

[(243)(105/16)(180)(5× 10−4)

]1/4

≈ 11.5; n = 12

15. (a) n >

[(π3)(1)(24)(10−3)

]1/2

≈ 35.9; n = 36 (b) n >

[(π3)(1)(12)(10−3)

]1/2

≈ 50.8; n = 51

(c) n >

[(π5)(1)

(180)(10−3)

]1/4

≈ 6.4; n = 8

16. (a) n >

[(1)(1)

(24)(10−3)

]1/2

≈ 6.5; n = 7 (b) n >

[(1)(1)

(12)(10−3)

]1/2

≈ 9.1; n = 10

(c) n >

[(1)(1)

(180)(10−3)

]1/4

≈ 1.5; n = 2

17. (a) n >

[(8)(e−1)(24)(10−6)

]1/2

≈ 350.2; n = 351 (b) n >

[(8)(e−1)(12)(10−6)

]1/2

≈ 495.2; n = 496

(c) n >

[(32)(e−1)(180)(10−6)

]1/4

≈ 15.99; n = 16

18. (a) n >

[(8)(8)

(24)(10−6)

]1/2

≈ 1632.99; n = 1633 (b) n >

[(8)(8)

(12)(10−6)

]1/2

≈ 2309.4;n = 2310

(c) n >

[(32)(384)(180)(10−6)

]1/4

≈ 90.9; n = 91

January 27, 2005 11:45 L24-CH08 Sheet number 40 Page number 376 black

376 Chapter 8

19. g(X0) = aX20 + bX0 + c = 4a+ 2b+ c = f(X0) = 1/X0 = 1/2; similarly

9a+ 3b+ c = 1/3, 16a+ 4b+ c = 1/4. Three equations in three unknowns, with solutiona = 1/24, b = −3/8, c = 13/12, g(x) = x2/24− 3x/8 + 13/12.∫ 4

2g(x) dx =

∫ 4

2

(x2

24− 3x8+1312

)dx =

2536

∆x3[f(X0) + 4f(X1) + f(X2)] =

13

[12+43+14

]=2536

20. Suppose g(x) = ax2 + bx + c passes through the points (0, f(0)) = (0, 0), (m, f(m)) = (1/6, 1/4),and (1/3, f(2m)) = (1/3, 3/4). Then g(0) = c = 0, 1/4 = g(1/6) = a/36 + b/6, and3/4 = g(1/3) = a/9 + b/3, with solution a = 9/2, b = 3/4 or g(x) = 9x2/2 + 3x/4. Then(∆x/3)(Y0 + 4Y1O + Y2) = (1/18)(0 + 4(1/4) + 3/4) = 7/72, and∫ 1/3

0g(x) dx =

[(3/2)x3 + (3/8)x2

]1/3

0= 1/18 + 1/24 = 7/72

21. 0.746824948,0.746824133

22. 1.236098366,1.236067977

23. 1.511518747,1.515927142

24. 2.418388347,2.418399152

25. 0.805376152,0.804776489

26. 1.536963087,1.544294774

27. (a) 3.142425985, |EM | ≈ 0.000833331(b) 3.139925989, |ET | ≈ 0.001666665(c) 3.141592614, |ES | ≈ 0.000000040

28. (a) 3.152411433, |EM | ≈ 0.010818779(b) 3.104518326, |ET | ≈ 0.037074328(c) 3.127008159, |ES | ≈ 0.014584495

29. S14 = 0.693147984, |ES | ≈ 0.000000803 = 8.03 × 10−7; the method used in Example 6 resultsin a value of n which ensures that the magnitude of the error will be less than 10−6, this is notnecessarily the smallest value of n.

30. (a) underestimates, because the graph of cosx2 is concave down on the interval (0, 1)(b) overestimates, because the graph of cosx2 is concave up on the interval (3/2, 2)

31. f(x) = x sinx, f ′′(x) = 2 cosx− x sinx, |f ′′(x)| ≤ 2| cosx|+ |x| | sinx| ≤ 2 + 2 = 4 so K2 ≤ 4,

n >

[(8)(4)

(24)(10−4)

]1/2

≈ 115.5; n = 116 (a smaller n might suffice)

32. f(x) = ecos x, f ′′(x) = (sin2 x)ecos x − (cosx)ecos x, |f ′′(x)| ≤ ecos x(sin2 x+ | cosx|) ≤ 2e so

K2 ≤ 2e, n >[(1)(2e)

(24)(10−4)

]1/2

≈ 47.6; n = 48 (a smaller n might suffice)

33. f(x) = x√x, f ′′(x) =

34√x, limx→0+

|f ′′(x)| = +∞

34. f(x) = sin√x, f ′′(x) = −

√x sin

√x+ cos

√x

4x3/2 , limx→0+

|f ′′(x)| = +∞

35. L =∫ π/2

−π/2

√1 + sin2 x dx ≈ 3.820187623 36. L =

∫ 3

1

√1 + 1/x4 dx ≈ 2.146822803

January 27, 2005 11:45 L24-CH08 Sheet number 41 Page number 377 black

Exercise Set 8.7 377

37.∫ 20

0v dt ≈ 15

(3)(6)[0 + 4(35.2) + 2(60.1) + 4(79.2) + 2(90.9) + 4(104.1) + 114.4] ≈ 1078 ft

38. 0

0

1

0.02

2

0.08

3

0.20

4

0.40

t

a

5

0.60

6

0.70

7

0.60

8

0

∫ 8

0a dt≈ 8

(3)(8)[0 + 4(0.02) + 2(0.08) + 4(0.20) + 2(0.40) + 4(0.60) + 2(0.70) + 4(0.60) + 0]

≈ 2.7 cm/s

39.∫ 180

0v dt ≈ 180

(3)(6)[0.00 + 4(0.03) + 2(0.08) + 4(0.16) + 2(0.27) + 4(0.42) + 0.65] = 37.9 mi

40.∫ 1800

0(1/v)dx ≈ 1800

(3)(6)

[13100

+42908

+22725

+42549

+22379

+42216

+12059

]≈ 0.71 s

41. V =∫ 16

0πr2dy = π

∫ 16

0r2dy≈ π 16

(3)(4)[(8.5)2 + 4(11.5)2 + 2(13.8)2 + 4(15.4)2 + (16.8)2]

≈ 9270 cm3 ≈ 9.3 L

42. A =∫ 600

0h dx ≈ 600

(3)(6)[0 + 4(7) + 2(16) + 4(24) + 2(25) + 4(16) + 0] = 9000 ft2,

V = 75A ≈ 75(9000) = 675, 000 ft3

43. (a) The maximum value of |f ′′(x)| is approximately 3.8442(b) n = 18(c) 0.9047406684

44. (a) The maximum value of |f ′′(x)| is approximately 1.46789(b) n = 8(c) 1.112830350

45. (a) The maximum value of |f (4)(x)| is approximately 42.5518.(b) n = 4(c) 0.9045241594

46. (a) The maximum value of |f (4)(x)| is approximately 7.02267.(b) n = 4(c) 1.111442702

47. (a) Left endpoint approximation ≈ b− an[y0 + y1 + . . .+ yn−2 + yn−1]

Right endpoint approximation ≈ b− an[y1 + y2 + . . .+ yn−1 + yn]

Average of the two =b− an

12[y0 + 2y1 + 2y2 + . . .+ 2yn−2 + 2yn−1 + yn]

January 27, 2005 11:45 L24-CH08 Sheet number 42 Page number 378 black

378 Chapter 8

(b) Area of trapezoid = (xk+1 − xk)yk + yk+1

2. If we sum from k = 0

to k = n− 1 then we get the right hand side of (2).y

x

1

x xk

yk

k+1

yk+1

2

3

48. right endpoint, trapezoidal, midpoint, left endpoint

49. Given g(x) = Ax2 +Bx+ C, suppose ∆x = 1 and m = 0. Then set Y0 = g(−1), Y1 = g(0),Y2 = g(1). Also Y0 = g(−1) = A − B + C, Y1 = g(0) = C, Y2 = g(1) = A + B + C, with solutionC = Y1, B = 1

2 (Y2 − Y0), and A = 12 (Y0 + Y2)− Y1.

Then∫ 1

−1g(x) dx = 2

∫ 1

0(Ax2 +C) dx =

23A+2C =

13(Y0 + Y2)−

23Y1 +2Y1 =

13(Y0 +4Y1 + Y2),

which is exactly what one gets applying the Simpson’s Rule.The general case with the interval (m −∆x,m + ∆x) and values Y0, Y1, Y2, can be converted by

the change of variables z =x−m∆x

. Set g(x) = h(z) = h((x−m)/∆x) to get dx = ∆x dz and

∆x∫ m+∆x

m−∆xh(z) dz =

∫ 1

−1g(x) dx. Finally, Y0 = g(m−∆x) = h(−1),

Y1 = g(m) = h(0), Y2 = g(m+∆x) = h(1).

50. From Exercise 49 we know, for i = 0, 1, . . . , n− 1 that∫ x2i+2

x2i

gi(x) dx =13b− a2n

[y2i + 4y2i+1 + y2i+2],

becauseb− a2n

is the width of the partition and acts as ∆x in Exercise 49.

Summing over all the subintervals note that y0 and y2n are only listed once; so∫ b

a

f(x) dx =n−1∑i=0

(∫ x2i+2

x2i

gi(x) dx)

=13b− a2n

[y0 + 4y1 + 2y2 + 4y3 + . . .+ 4y2n−2 + 2y2n−1 + y2n]

EXERCISE SET 8.8

1. (a) improper; infinite discontinuity at x = 3(b) continuous integrand, not improper(c) improper; infinite discontinuity at x = 0(d) improper; infinite interval of integration(e) improper; infinite interval of integration and infinite discontinuity at x = 1(f) continuous integrand, not improper

2. (a) improper if p > 0 (b) improper if 1 ≤ p ≤ 2(c) integrand is continuous for all p, not improper

January 27, 2005 11:45 L24-CH08 Sheet number 43 Page number 379 black

Exercise Set 8.8 379

3. lim�→+∞

(−12e−2x

)]�0=12lim

�→+∞(−e−2� + 1) =

12

4. lim�→+∞

12ln(1 + x2)

]�−1= lim�→+∞

12[ln(1 + .2)− ln 2] = +∞, divergent

5. lim�→+∞

−2 coth−1 x

]�3= lim�→+∞

(2 coth−1 3− 2 coth−1 .

)= 2 coth−1 3

6. lim�→+∞

−12e−x

2]�

0= lim�→+∞

12

(−e−�2 + 1

)= 1/2

7. lim�→+∞

− 12 ln2 x

]�e

= lim�→+∞

[− 12 ln2 .

+12

]=12

8. lim�→+∞

2√lnx]�

2= lim�→+∞

(2√ln .− 2

√ln 2) = +∞, divergent

9. lim�→−∞

− 14(2x− 1)2

]0

= lim�→−∞

14[−1 + 1/(2.− 1)2] = −1/4

10. lim�→−∞

13tan−1 x

3

]3

= lim�→−∞

13

4− tan−1 .

3

]=13[π/4− (−π/2)] = π/4

11. lim�→−∞

13e3x]0

= lim�→−∞

[13− 13e3�]=13

12. lim�→−∞

−12ln(3− 2ex)

]0

= lim�→−∞

12ln(3− 2e�) = 1

2ln 3

13.∫ +∞

−∞x dx converges if

∫ 0

−∞x dx and

∫ +∞

0x dx both converge; it diverges if either (or both)

diverges.∫ +∞

0x dx = lim

�→+∞

12x2]�

0= lim�→+∞

12.2 = +∞ so

∫ +∞

−∞x dx is divergent.

14.∫ +∞

0

x√x2 + 2

dx = lim�→+∞

√x2 + 2

]�0= lim�→+∞

(√.2 + 2−

√2) = +∞

so∫ ∞−∞

x√x2 + 2

dx is divergent.

15.∫ +∞

0

x

(x2 + 3)2dx = lim

�→+∞− 12(x2 + 3)

]�0= lim�→+∞

12[−1/(.2 + 3) + 1/3] = 1

6,

similarly∫ 0

−∞

x

(x2 + 3)2dx = −1/6 so

∫ ∞−∞

x

(x2 + 3)2dx = 1/6 + (−1/6) = 0

January 27, 2005 11:45 L24-CH08 Sheet number 44 Page number 380 black

380 Chapter 8

16.∫ +∞

0

e−t

1 + e−2t dt = lim�→+∞

− tan−1(e−t)]�

0= lim�→+∞

[− tan−1(e−�) +

π

4

]=π

4,

∫ 0

−∞

e−t

1 + e−2t dt = lim�→−∞

− tan−1(e−t)]0

= lim�→−∞

[−π4+ tan−1(e−�)

]=π

4,

∫ +∞

−∞

e−t

1 + e−2t dt =π

4+π

4=π

2

17. lim�→4−

− 1x− 4

]�0= lim�→4−

[− 1.− 4 −

14

]= +∞, divergent

18. lim�→0+

32x2/3

]8

= lim�→0+

32(4− .2/3) = 6

19. lim�→π/2−

− ln(cosx)]�

0= lim�→π/2−

− ln(cos .) = +∞, divergent

20. lim�→4−

−2√4− x

]�0= lim�→4−

2(−√4− .+ 2) = 4

21. lim�→1−

sin−1 x

]�0= lim�→1−

sin−1 . = π/2

22. lim�→−3+

−√9− x2

]1

= lim�→−3+

(−√8 +

√9− .2) = −

√8

23. lim�→π/3+

√1− 2 cosx

]π/2�

= lim�→π/3+

(1−√1− 2 cos .) = 1

24. lim�→π/4−

− ln(1− tanx)]�

0= lim�→π/4−

− ln(1− tan .) = +∞, divergent

25.∫ 2

0

dx

x− 2 = lim�→2−

ln |x− 2|]�

0= lim�→2−

(ln |.− 2| − ln 2) = −∞, divergent

26.∫ 2

0

dx

x2 = lim�→0+

−1/x]2

= lim�→0+

(−1/2 + 1/.) = +∞ so∫ 2

−2

dx

x2 is divergent

27.∫ 8

0x−1/3dx = lim

�→0+

32x2/3

]8

= lim�→0+

32(4− .2/3) = 6,

∫ 0

−1x−1/3dx = lim

�→0−

32x2/3

]�−1= lim�→0−

32(.2/3 − 1) = −3/2

so∫ 8

−1x−1/3dx = 6 + (−3/2) = 9/2

28.∫ 1

0

dx

(x− 1)2/3 = lim�→1−

3(x− 1)1/3]�

0= lim�→1−

3[(.− 1)1/3 − (−1)1/3] = 3

January 27, 2005 11:45 L24-CH08 Sheet number 45 Page number 381 black

Exercise Set 8.8 381

29. Define∫ +∞

0

1x2 dx =

∫ a

0

1x2 dx+

∫ +∞

a

1x2 dx where a > 0; take a = 1 for convenience,∫ 1

0

1x2 dx = lim

�→0+(−1/x)

]1

= lim�→0+

(1/.− 1) = +∞ so∫ +∞

0

1x2 dx is divergent.

30. Define∫ +∞

1

dx

x√x2 − 1

=∫ a

1

dx

x√x2 − 1

+∫ +∞

a

dx

x√x2 − 1

where a > 1,

take a = 2 for convenience to get∫ 2

1

dx

x√x2 − 1

= lim�→1+

sec−1 x

]2

= lim�→1+

(π/3− sec−1 .) = π/3,

∫ +∞

2

dx

x√x2 − 1

= lim�→+∞

sec−1 x

]�2= π/2− π/3 so

∫ +∞

1

dx

x√x2 − 1

= π/2.

31.∫ +∞

0

e−√x

√xdx = 2

∫ +∞

0e−udu = 2 lim

�→+∞

(−e−u

)]�0= 2 lim

�→+∞

(1− e−�

)= 2

32.∫ +∞

12

dx√x(x+ 4)

= 2∫ +∞

2√

3

du

u2 + 4= 2 lim

�→+∞

12tan−1 u

2

]�2√

3= lim�→+∞

tan−1 .

2− tan−1

√3 =

π

6

33.∫ +∞

0

e−x√1− e−x

dx =∫ 1

0

du√u= lim

�→0+2√u

]1

= lim�→0+

2(1−√.) = 2

34.∫ +∞

0

e−x√1− e−2x

dx = −∫ 0

1

du√1− u2

=∫ 1

0

du√1− u2

= lim�→1

sin−1 u

]�0= lim�→1

sin−1 . =π

2

35. lim�→+∞

∫ �

0e−x cosx dx = lim

�→+∞

12e−x(sinx− cosx)

]�0= 1/2

36. A =∫ +∞

0xe−3xdx = lim

�→+∞−19(3x+ 1)e−3x

]�0= 1/3

37. (a) 2.726585 (b) 2.804364 (c) 0.219384 (d) 0.504067

39. 1 +(dy

dx

)2

= 1 +4− x2/3

x2/3 =4x2/3 ; the arc length is

∫ 8

0

2x1/3 dx = 3x

2/3

∣∣∣∣∣8

0

= 12

40. 1 +(dy

dx

)2= 1 +

x2

4− x2 =4

4− x2 ; the arc length is

∫ 2

0

√4

4− x2 dx = lim�→2−

∫ �

0

2√4− x2

dx = lim�→2−

2 sin−1 x

2

]�0= 2 sin−1 1 = π

41.∫lnx dx = x lnx− x+ C,

∫ 1

0lnx dx = lim

�→0+

∫ 1

lnx dx = lim�→0+

(x lnx− x)]1

= lim�→0+

(−1− . ln .+ .),

but lim�→0+

. ln . = lim�→0+

ln .1/.

= lim�→0+

(−.) = 0 so∫ 1

0lnx dx = −1

January 27, 2005 11:45 L24-CH08 Sheet number 46 Page number 382 black

382 Chapter 8

42.∫lnxx2 dx = −

lnxx− 1x+ C,

∫ +∞

1

lnxx2 dx = lim

�→+∞

∫ �

1

lnxx2 dx = lim

�→+∞

(− lnxx− 1x

)]�1= lim�→+∞

(− ln ..− 1.+ 1),

but lim�→+∞

ln ..= lim�→+∞

1.= 0 so

∫ +∞

1

lnxx2 = 1

43.∫ +∞

0e−3xdx = lim

�→+∞

∫ �

0e−3xdx = lim

�→+∞

(−13e−3x

)]�0= lim�→+∞

(−13e−3� +

19

)=13

44. A =∫ +∞

4

8x2 − 4dx = lim

�→+∞2 ln

x− 2x+ 2

]�4= lim�→+∞

2[ln.− 2.+ 2

− ln 13

]= 2 ln 3

45. (a) V = π∫ +∞

0e−2xdx = −π

2lim

�→+∞e−2x

]�0= π/2

(b) S = 2π∫ +∞

0e−x

√1 + e−2xdx, let u = e−x to get

S = −2π∫ 0

1

√1 + u2du = 2π

[u

2

√1 + u2 +

12ln∣∣∣u+√1 + u2

∣∣∣]1

0= π

[√2 + ln(1 +

√2)]

47. (a) For x ≥ 1, x2 ≥ x, e−x2 ≤ e−x

(b)∫ +∞

1e−x dx = lim

�→+∞

∫ �

1e−x dx = lim

�→+∞−e−x

]�1= lim�→+∞

(e−1 − e−�) = 1/e

(c) By Parts (a) and (b) and Exercise 46(b),∫ +∞

1e−x

2dx is convergent and is ≤ 1/e.

48. (a) If x ≥ 0 then ex ≥ 1, 12x+ 1

≤ ex

2x+ 1

(b) lim�→+∞

∫ �

0

dx

2x+ 1= lim�→+∞

12ln(2x+ 1)

]�0= +∞

(c) By Parts (a) and (b) and Exercise 46(a),∫ +∞

0

ex

2x+ 1dx is divergent.

49. V = lim�→+∞

∫ �

1(π/x2) dx = lim

�→+∞−(π/x)

]�1= lim�→+∞

(π − π/.) = π

A = lim�→+∞

∫ �

12π(1/x)

√1 + 1/x4 dx;

use Exercise 46(a) with f(x) = 2π/x, g(x) = (2π/x)√1 + 1/x4

and a = 1 to see that the area is infinite.

50. (a) 1 ≤√x3 + 1x

for x ≥ 2,∫ +∞

21dx = +∞

(b)∫ +∞

2

x

x5 + 1dx ≤

∫ +∞

2

dx

x4 = lim�→+∞

− 13x3

]�2= 1/24

(c)∫ ∞

0

xex

2x+ 1dx ≥

∫ +∞

1

xex

2x+ 1≥∫ +∞

1

dx

2x+ 1= +∞

January 27, 2005 11:45 L24-CH08 Sheet number 47 Page number 383 black

Exercise Set 8.8 383

51. The area under the curve y =1

1 + x2 , above the x-axis, and to the

right of the y-axis is given by∫ ∞

0

11 + x2 . Solving for

y =√

1−yy , the area is also given by the improper integral

∫ 1

0

√1− yy

dy.

0 42

x

52. (b) u =√x,

∫ +∞

0

cos√x√xdx = 2

∫ +∞

0cosu du;

∫ +∞

0cosu du diverges by Part (a).

53. Let x = r tan θ to get∫

dx

(r2 + x2)3/2=1r2

∫cos θ dθ =

1r2 sin θ + C =

x

r2√r2 + x2

+ C

so u =2πNIrk

lim�→+∞

x

r2√r2 + x2

]�a

=2πNIkr

lim�→+∞

(./√r2 + .2 − a/

√r2 + a2)

=2πNIkr

(1− a/√r2 + a2).

54. Let a2 =M

2RTto get

(a) v =4√π

(M

2RT

)3/2 12

(M

2RT

)−2

=2√π

√2RTM

=

√8RTπM

(b) v2rms =

4√π

(M

2RT

)3/2 3√π8

(M

2RT

)−5/2

=3RTM

so vrms =

√3RTM

55. (a) Satellite’s weight = w(x) = k/x2 lb when x = distance from center of Earth; w(4000) = 6000

so k = 9.6× 1010 and W =∫ 4000+�

40009.6× 1010x−2dx mi·lb.

(b)∫ +∞

40009.6× 1010x−2dx = lim

�→+∞−9.6× 1010/x

]�4000

= 2.4× 107 mi·lb

56. (a) L{1} =∫ +∞

0e−stdt = lim

�→+∞−1se−st

]�0=1s

(b) L{e2t} =∫ +∞

0e−ste2tdt =

∫ +∞

0e−(s−2)tdt = lim

�→+∞− 1s− 2e

−(s−2)t]�

0=

1s− 2

(c) L{sin t} =∫ +∞

0e−st sin t dt = lim

�→+∞

e−st

s2 + 1(−s sin t− cos t)

]�0=

1s2 + 1

(d) L{cos t} =∫ +∞

0e−st cos t dt = lim

�→+∞

e−st

s2 + 1(−s cos t+ sin t)

]�0=

s

s2 + 1

January 27, 2005 11:45 L24-CH08 Sheet number 48 Page number 384 black

384 Chapter 8

57. (a) L{f(t)} =∫ +∞

0te−st dt = lim

�→+∞−(t/s+ 1/s2)e−st

]�0=1s2

(b) L{f(t)} =∫ +∞

0t2e−st dt = lim

�→+∞−(t2/s+ 2t/s2 + 2/s3)e−st

]�0=2s3

(c) L{f(t)} =∫ +∞

3e−stdt = lim

�→+∞−1se−st

]�3=e−3s

s

58. 10 100 1000 10,000

0.8862269 0.8862269 0.8862269 0.8862269

59. (a) u =√ax, du =

√a dx, 2

∫ +∞

0e−ax

2dx =

2√a

∫ +∞

0e−u

2du =

√π/a

(b) x =√2σu, dx =

√2σ du,

2√2πσ

∫ +∞

0e−x

2/2σ2dx =

2√π

∫ +∞

0e−u

2du = 1

60. (a)∫ 3

0e−x

2dx ≈ 0.8862;

√π/2 ≈ 0.8862

(b)∫ +∞

0e−x

2dx =

∫ 3

0e−x

2dx+

∫ +∞

3e−x

2dx so E =

∫ +∞

3e−x

2dx <

∫ +∞

3xe−x

2dx =

12e−9 < 7× 10−5

61. (a)∫ 4

0

1x6 + 1

dx ≈ 1.047; π/3 ≈ 1.047

(b)∫ +∞

0

1x6 + 1

dx =∫ 4

0

1x6 + 1

dx+∫ +∞

4

1x6 + 1

dx so

E =∫ +∞

4

1x6 + 1

dx <

∫ +∞

4

1x6 dx =

15(4)5

< 2× 10−4

62. If p = 0, then∫ +∞

0(1)dx = lim

�→+∞x

]�0= +∞,

if p �= 0, then∫ +∞

0epxdx = lim

�→+∞

1pepx]�

0= lim�→+∞

1p(ep� − 1) =

{−1/p, p < 0+∞, p > 0 .

63. If p = 1, then∫ 1

0

dx

x= lim

�→0+lnx]1

= +∞;

if p �= 1, then∫ 1

0

dx

xp= lim

�→0+

x1−p

1− p

]1

= lim�→0+

[(1− .1−p)/(1− p)] ={1/(1− p), p < 1+∞, p > 1 .

64. u =√1− x, u2 = 1− x, 2u du = −dx;

−2∫ 0

1

√2− u2 du = 2

∫ 1

0

√2− u2 du =

[u√2− u2 + 2 sin−1(u/

√2)]1

0=√2 + π/2

65. 2∫ 1

0cos(u2)du ≈ 1.809

66. −2∫ 0

1sin(1− u2)du = 2

∫ 1

0sin(1− u2)du ≈ 1.187

January 27, 2005 11:45 L24-CH08 Sheet number 49 Page number 385 black

Review Exercises, Chapter 8 385

67. (a) Γ(1) =∫ +∞

0e−tdt = lim

�→+∞−e−t

]�0= lim�→+∞

(−e−� + 1) = 1

(b) Γ(x+ 1) =∫ +∞

0txe−tdt; let u = tx, dv = e−tdt to get

Γ(x+ 1) = −txe−t]+∞

0+ x

∫ +∞

0tx−1e−tdt = −txe−t

]+∞

0+ xΓ(x)

limt→+∞

txe−t = limt→+∞

tx

et= 0 (by multiple applications of L’Hopital’s rule)

so Γ(x+ 1) = xΓ(x)

(c) Γ(2) = (1)Γ(1) = (1)(1) = 1, Γ(3) = 2Γ(2) = (2)(1) = 2, Γ(4) = 3Γ(3) = (3)(2) = 6

It appears that Γ(n) = (n− 1)! if n is a positive integer.

(d) Γ(12

)=∫ +∞

0t−1/2e−tdt = 2

∫ +∞

0e−u

2du (with u =

√t) = 2(

√π/2) =

√π

(e) Γ(32

)=12Γ(12

)=12√π, Γ

(52

)=32Γ(32

)=34√π

68. (a) t = − lnx, x = e−t, dx = −e−tdt,∫ 1

0(lnx)ndx = −

∫ 0

+∞(−t)ne−tdt = (−1)n

∫ +∞

0tne−tdt = (−1)nΓ(n+ 1)

(b) t = xn, x = t1/n, dx = (1/n)t1/n−1dt,∫ +∞

0e−x

n

dx = (1/n)∫ +∞

0t1/n−1e−tdt = (1/n)Γ(1/n) = Γ(1/n+ 1)

69. (a)√cos θ − cos θ0 =

√2[sin2(θ0/2)− sin2(θ/2)

]=√2(k2 − k2 sin2 φ) =

√2k2 cos2 φ

=√2 k cosφ; k sinφ = sin(θ/2) so k cosφdφ =

12cos(θ/2) dθ =

12

√1− sin2(θ/2) dθ

=12

√1− k2 sin2 φdθ, thus dθ =

2k cosφ√1− k2 sin2 φ

dφ and hence

T =

√8Lg

∫ π/2

0

1√2k cosφ

· 2k cosφ√1− k2 sin2 φ

dφ = 4

√L

g

∫ π/2

0

1√1− k2 sin2 φ

(b) If L = 1.5 ft and θ0 = (π/180)(20) = π/9, then

T =√32

∫ π/2

0

dφ√1− sin2(π/18) sin2 φ

≈ 1.37 s.

REVIEW EXERCISES, CHAPTER 8

1. u = 4 + 9x, du = 9 dx,19

∫u1/2 du =

227(4 + 9x)3/2 + C

2. u = πx, du = π dx,1π

∫cosu du =

1πsinu+ C =

1πsinπx+ C

January 27, 2005 11:45 L24-CH08 Sheet number 50 Page number 386 black

386 Chapter 8

3. u = cos θ, −∫u1/2du = −2

3cos3/2 θ + C

4. u = lnx, du =dx

x,

∫du

u= ln |u|+ C = ln |lnx|+ C

5. u = tan(x2),12

∫u2du =

16tan3(x2) + C

6. u =√x, x = u2, dx = 2u du,

2∫ 3

0

u2

u2 + 9du = 2

∫ 3

0

(1− 9

u2 + 9

)du =

(2u− 6 tan−1 u

3

)]3

0= 6− 3

7. (a) With u =√x:∫

1√x√2− x

dx = 2∫

1√2− u2

du = 2 sin−1(u/√2) + C = 2 sin−1(

√x/2) + C;

with u =√2− x :∫

1√x√2− x

dx = −2∫

1√2− u2

du = −2 sin−1(u/√2) + C = −2 sin−1(

√2− x/

√2) + C1;

completing the square:∫1√

1− (x− 1)2dx = sin−1(x− 1) + C.

(b) In the three results in Part (a) the antiderivatives differ by a constant, in particular2 sin−1(

√x/2) = π − 2 sin−1(

√2− x/

√2) = π/2 + sin−1(x− 1).

8. (a) u = x2, dv =x√x2 + 1

dx, du = 2x dx, v =√x2 + 1;

∫ 1

0

x3√x2 + 1

dx = x2√x2 + 1

]10− 2

∫ 1

0x(x2 + 1)1/2dx

=√2− 2

3(x2 + 1)3/2

]1

0=√2− 2

3[2√2− 1] = (2−

√2)/3

(b) u2 = x2 + 1, x2 = u2 − 1, 2x dx = 2u du, x dx = u du;∫ 1

0

x3√x2 + 1

dx =∫ 1

0

x2√x2 + 1

x dx=∫ √2

1

u2 − 1u

u du

=∫ √2

1(u2 − 1)du =

(13u3 − u

)]√2

1= (2−

√2)/3

9. u = x, dv = e−xdx, du = dx, v = −e−x;∫xe−xdx = −xe−x +

∫e−xdx = −xe−x − e−x + C

10. u = x, dv = sin 2x dx, du = dx, v = −12cos 2x;∫

x sin 2x dx = −12x cos 2x+

12

∫cos 2x dx = −1

2x cos 2x+

14sin 2x+ C

January 27, 2005 11:45 L24-CH08 Sheet number 51 Page number 387 black

Review Exercises, Chapter 8 387

11. u = ln(2x+ 3), dv = dx, du =2

2x+ 3dx, v = x;

∫ln(2x+ 3)dx = x ln(2x+ 3)−

∫2x

2x+ 3dx

but∫

2x2x+ 3

dx =∫ (

1− 32x+ 3

)dx = x− 3

2ln(2x+ 3) + C1 so∫

ln(2x+ 3)dx = x ln(2x+ 3)− x+ 32ln(2x+ 3) + C

12. u = tan−1(2x), dv = dx, du =2

1 + 4x2 dx, v = x;∫tan−1(2x)dx = x tan−1(2x)−

∫2x

1 + 4x2 dx = x tan−1(2x)− 1

4ln(1 + 4x2) + C

13. Repeated RepeatedDifferentiation Antidifferentiation

8x4 cos 2x+

32x3 12sin 2x

−96x2 −1

4cos 2x

+

192x −18sin 2x

−192

116cos 2x

+

0132sin 2x

∫8x4 cos 2x dx = (4x4 − 12x2 + 6) sin 2x+ (8x3 − 12x) cos 2x+ C

14. distance =∫ 5

0t2e−tdt;u = t2, dv = e−tdt, du = 2tdt, v = −e−t,

distance = −t2e−t]5

0+ 2

∫ 5

0te−tdt;u = 2t, dv = e−tdt, du = 2dt, v = −e−t,

distance = −25e−5 − 2te−t]5

0+ 2

∫ 5

0e−tdt = −25e−5 − 10e−5 − 2e−t

]50

= −25e−5 − 10e−5 − 2e−5 + 2 = −37e−5 + 2

15.∫sin2 5θ dθ =

12

∫(1− cos 10θ)dθ = 1

2θ − 1

20sin 10θ + C

16.∫sin3 2x cos2 2x dx=

∫(1− cos2 2x) cos2 2x sin 2x dx

=∫(cos2 2x− cos4 2x) sin 2x dx = −1

6cos3 2x+

110cos5 2x+ C

17.∫sinx cos 2x dx =

12

∫(sin 3x− sinx)dx = −1

6cos 3x+

12cosx+ C

January 27, 2005 11:45 L24-CH08 Sheet number 52 Page number 388 black

388 Chapter 8

18.∫ π/6

0sin 2x cos 4x dx=

12

∫ π/6

0(sin 6x− sin 2x)dx =

[− 112cos 6x+

14cos 2x

]π/60

= [(−1/12)(−1) + (1/4)(1/2)]− [−1/12 + 1/4] = 1/24

19. u = 2x,∫sin4 2x dx=

12

∫sin4 u du =

12

[−14sin3 u cosu+

34

∫sin2 u du

]

= −18sin3 u cosu+

38

[−12sinu cosu+

12

∫du

]

= −18sin3 u cosu− 3

16sinu cosu+

316u+ C

= −18sin3 2x cos 2x− 3

16sin 2x cos 2x+

38x+ C

20. u = x2,∫x cos5(x2)dx=

12

∫cos5 u du =

12

∫(cosu)(1− sin2 u)2 du

=12

∫cosu du−

∫cosu sin2 u du+

12

∫cosu sin4 u du

=12sinu− 1

3sin3 u+

110sin5 u+ C

=12sin(x2)− 1

3sin3(x2) +

110sin5(x2) + C

21. x = 3 sin θ, dx = 3 cos θ dθ,

9∫sin2 θ dθ=

92

∫(1− cos 2θ)dθ = 9

2θ − 9

4sin 2θ + C =

92θ − 9

2sin θ cos θ + C

=92sin−1(x/3)− 1

2x√9− x2 + C

22. x = 4 sin θ, dx = 4 cos θ dθ,

116

∫1

sin2 θdθ =

116

∫csc2 θ dθ = − 1

16cot θ + C = −

√16− x2

16x+ C

23. x = sec θ, dx = sec θ tan θ dθ,∫sec θ dθ = ln | sec θ + tan θ|+ C = ln

∣∣∣x+√x2 − 1∣∣∣+ C

24. x = 5 sec θ, dx = 5 sec θ tan θ dθ,

25∫sec3 θ dθ=

252sec θ tan θ +

252ln | sec θ + tan θ|+ C1

=12x√x2 − 25 + 25

2ln |x+

√x2 − 25|+ C

January 27, 2005 11:45 L24-CH08 Sheet number 53 Page number 389 black

Review Exercises, Chapter 8 389

25. x = 3 tan θ, dx = 3 sec2 θ dθ,

9∫tan2 θ sec θ dθ= 9

∫sec3 θ dθ − 9

∫sec θ dθ

=92sec θ tan θ − 9

2ln | sec θ + tan θ|+ C

=12x√9 + x2 − 9

2ln |13

√9 + x2 +

13x|+ C

26. 2x = tan θ, 2 dx = sec2 θ dθ,∫sec2 θ csc θ dθ=

∫(sec θ tan θ + csc θ) dθ

= sec θ − ln | csc θ + cot θ|+ C

=√1 + 4x2 − ln

∣∣∣∣∣√1 + 4x2

2x+12x

∣∣∣∣∣+ C

27.1

(x+ 4)(x− 1) =A

x+ 4+

B

x− 1 ; A = −15, B =

15so

−15

∫1

x+ 4dx+

15

∫1

x− 1dx = −15ln |x+ 4|+ 1

5ln |x− 1|+ C = 1

5ln∣∣∣∣x− 1x+ 4

∣∣∣∣+ C

28.1

(x+ 1)(x+ 7)=

A

x+ 1+

B

x+ 7; A =

16, B = −1

6so

16

∫1

x+ 1dx− 1

6

∫1

x+ 7dx =

16ln |x+ 1| − 1

6ln |x+ 7|+ C = 1

6ln∣∣∣∣x+ 1x+ 7

∣∣∣∣+ C29.

x2 + 2x+ 2

= x− 2 + 6x+ 2

,∫ (

x− 2 + 6x+ 2

)dx =

12x2 − 2x+ 6 ln |x+ 2|+ C

30.x2 + x− 16(x+ 1)(x− 3)2 =

A

x+ 1+

B

x− 3 +C

(x− 3)2 ; A = −1, B = 2, C = −1 so

−∫

1x+ 1

dx+ 2∫

1x− 3 dx−

∫1

(x− 3)2 dx

= − ln |x+ 1|+ 2 ln |x− 3|+ 1x− 3 + C = ln

(x− 3)2|x+ 1| +

1x− 3 + C

31.x2

(x+ 2)3=

A

x+ 2+

B

(x+ 2)2+

C

(x+ 2)3; A = 1, B = −4, C = 4 so

∫1

x+ 2dx− 4

∫1

(x+ 2)2dx+ 4

∫1

(x+ 2)3dx = ln |x+ 2|+ 4

x+ 2− 2(x+ 2)2

+ C

32.1

x(x2 + 1)=A

x+Bx+ Cx2 + 1

; A = 1, B = −1, C = 0 so

∫1

x3 + xdx = ln |x| − 1

2ln(x2 + 1) + C =

12ln

x2

x2 + 1+ C

January 27, 2005 11:45 L24-CH08 Sheet number 54 Page number 390 black

390 Chapter 8

33. (a) With x = sec θ:∫1

x3 − xdx =∫cot θ dθ = ln | sin θ|+ C = ln

√x2 − 1|x| + C; valid for |x| > 1.

(b) With x = sin θ:∫1

x3 − xdx = −∫

1sin θ cos θ

dθ = −∫2 csc 2θ dθ

= − ln | csc 2θ − cot 2θ|+ C = ln | cot θ|+ C = ln√1− x2

|x| + C, 0 < |x| < 1.

34. A =∫ 2

1

3− xx3 + x2 dx,

3− xx2(x+ 1)

=A

x+B

x2 +C

x+ 1; A = −4, B = 3, C = 4

A=[−4 ln |x| − 3

x+ 4 ln |x+ 1|

]2

1

= (−4 ln 2− 32+ 4 ln 3)− (−4 ln 1− 3 + 4 ln 2) = 3

2− 8 ln 2 + 4 ln 3 = 3

2+ 4 ln

34

35. #40 36. #52 37. #113

38. #108 39. #28 40. #71

41. exact value = 14/3 ≈ 4.666666667(a) 4.667600663, |EM | ≈ 0.000933996(b) 4.664795679, |ET | ≈ 0.001870988(c) 4.666651630, |ES | ≈ 0.000015037

42. exact value =12ln 5 ≈ 0.804718956

(a) 0.801605339, |EM | ≈ 0.003113617(b) 0.811019505, |ET | ≈ 0.006300549(c) 0.805041497, |ES | ≈ 0.000322541

43. f(x) =√x+ 1, f ′′(x) = −1

4(x+ 1)−3/2, f (4)(x) = −15

16(x+ 1)−7/2; K2 = 1/4, K4 = 15/16

(a) |EM | ≤272400

(1/4) = 0.002812500 (b) |ET | ≤271200

(1/4) = 0.005625000

(c) |ES | ≤243

180× 104 (15/16) ≈ 0.000126563

44. f(x) = 1/(2x+ 3), f ′′(x) = 8(2x+ 3)−3, f (4)(x) = 384(2x+ 3)−5; K2 = 8, K4 = 384

(a) |EM | ≤82400

(8) ≈ 0.026666667 (b) |ET | ≤81200

(8) ≈ 0.053333333

(c) |ES | ≤32

180× 104 (384) ≈ 0.006826667

45. (a) n >

[(27)(1/4)

(24)(5× 10−4)

]1/2

≈ 23.7; n = 24 (b) n >

[(27)(1/4)

(12)(5× 10−4)

]1/2≈ 33.5; n=34

(c) n >

[(243)(15/16)(180)(5× 10−4)

]1/4

≈ 7.1; n = 8

January 27, 2005 11:45 L24-CH08 Sheet number 55 Page number 391 black

Review Exercises, Chapter 8 391

46. (a) n >

[(8)(8)

(24)(10−6)

]1/2

≈ 1632.99; n = 1633 (b) n >

[(8)(8)

(12)(10−6)

]1/2

≈ 2309.4;n = 2310

(c) n >

[(32)(384)(180)(10−6)

]1/4

≈ 90.9; n = 92

47. lim�→+∞

(−e−x)]�

0= lim�→+∞

(−e−� + 1) = 1

48. lim�→−∞

12tan−1 x

2

]2

= lim�→−∞

12

4− tan−1 .

2

]=12[π/4− (−π/2)] = 3π/8

49. lim�→9−

−2√9− x

]�0= lim�→9−

2(−√9− .+ 3) = 6

50.∫ 1

0

12x− 1 dx =

∫ 1/2

0

12x− 1 dx+

∫ 1

1/2

12x− 1 dx = lim

�→1/2−

12ln(2x− 1) + lim

�→1/2+

12ln(2x− 1) +C

neither limit exists hence the integral diverges

51. A =∫ +∞

e

lnx− 1x2 dx = lim

�→+∞c− lnx

x

]�e

= 1/e

52. V = 2π∫ +∞

0xe−xdx = 2π lim

�→+∞−e−x(x+ 1)

]�0= 2π lim

�→+∞

[1− e−�(.+ 1)

]

but lim�→+∞

e−�(.+ 1) = lim�→+∞

.+ 1e�

= lim�→+∞

1e�= 0 so V = 2π

53.∫ +∞

0

dx

x2 + a2 = lim�→+∞

1atan−1(x/a)

]�0= lim�→+∞

1atan−1(./a) =

π

2a= 1, a = π/2

54. (a) integration by parts, u = x, dv = sinx dx (b) u-substitution: u = sinx(c) reduction formula (d) u-substitution: u = tanx(e) u-substitution: u = x3 + 1 (f) u-substitution: u = x+ 1(g) integration by parts: dv = dx, u = tan−1 x (h) trigonometric substitution: x = 2 sin θ(i) u-substitution: u = 4− x2

55. x =√3 tan θ, dx =

√3 sec2 θ dθ,

13

∫1sec θ

dθ =13

∫cos θ dθ =

13sin θ + C =

x

3√3 + x2

+ C

56. u = x, dv = cos 3x dx, du = dx, v =13sin 3x;∫

x cos 3x dx =13x sin 3x− 1

3

∫sin 3x dx =

13x sin 3x+

19cos 3x+ C

57. Use Endpaper Formula (31) to get∫tan7 θ dθ =

16tan6 θ − 1

4tan4 θ +

12tan2 θ + ln | cos θ|+ C.

58.∫

cos θ(sin θ − 3)2 + 3dθ, let u = sin θ − 3,

∫1

u2 + 3du =

1√3f tan−1[(sin θ − 3)/

√3] + C

January 27, 2005 11:45 L24-CH08 Sheet number 56 Page number 392 black

392 Chapter 8

59.∫sin2 2t cos3 2t dt=

∫sin2 2t(1− sin2 2t) cos 2t dt =

∫(sin2 2t− sin4 2t) cos 2t dt

=16sin3 2t− 1

10sin5 2t+ C

60.∫ 3

0

1(x− 3)2 dx = lim

�→3−

∫ �

0

1(x− 3)2 dx = lim

�→3−− 1x− 3

]�0which is clearly divergent,

so the integral diverges.

61. u = e2x, dv = cos 3x dx, du = 2e2xdx, v =13sin 3x;∫

e2x cos 3x dx =13e2x sin 3x− 2

3

∫e2x sin 3x dx. Use u = e2x, dv = sin 3x dx to get

∫e2x sin 3x dx = −1

3e2x cos 3x+

23

∫e2x cos 3x dx so

∫e2x cos 3x dx =

13e2x sin 3x+

29e2x cos 3x− 4

9

∫e2x cos 3x dx,

139

∫e2x cos 3x dx =

19e2x(3 sin 3x+2 cos 3x)+C1,

∫e2x cos 3x dx =

113e2x(3 sin 3x+2 cos 3x)+C

62. x = (1/√2) sin θ, dx = (1/

√2) cos θ dθ,

1√2

∫ π/2

−π/2cos4 θ dθ =

1√2

{14cos3 θ sin θ

]π/2−π/2

+34

∫ π/2

−π/2cos2 θ dθ

}

=34√2

{12cos θ sin θ

]π/2−π/2

+12

∫ π/2

−π/2dθ

}=

34√212π =

3π8√2

63.1

(x− 1)(x+ 2)(x− 3) =A

x− 1 +B

x+ 2+

C

x− 3 ; A = −16, B =

115, C =

110so

−16

∫1

x− 1dx+115

∫1

x+ 2dx+

110

∫1

x− 3dx

= −16ln |x− 1|+ 1

15ln |x+ 2|+ 1

10ln |x− 3|+ C

64. x =23sin θ, dx =

23cos θ dθ,

124

∫ π/6

0

1cos3 θ

dθ=124

∫ π/6

0sec3 θ dθ =

[148sec θ tan θ +

148ln | sec θ + tan θ|

]π/60

=148[(2/√3)(1/

√3) + ln |2/

√3 + 1/

√3|] = 1

48

(23+12ln 3)

65. u =√x− 4, x = u2 + 4, dx = 2u du,∫ 2

0

2u2

u2 + 4du = 2

∫ 2

0

[1− 4

u2 + 4

]du =

[2u− 4 tan−1(u/2)

]2

0= 4− π

January 27, 2005 11:45 L24-CH08 Sheet number 57 Page number 393 black

Review Exercises, Chapter 8 393

66. u =√ex − 1, ex = u2 + 1, x = ln(u2 + 1), dx =

2uu2 + 1

du,

∫ 1

0

2u2

u2 + 1du = 2

∫ 1

0

(1− 1

u2 + 1

)du = (2u− 2 tan−1 u)

]1

0= 2− π

2

67. u =√ex + 1, ex = u2 − 1, x = ln(u2 − 1), dx = 2u

u2 − 1du,∫2

u2 − 1du =∫ [

1u− 1 −

1u+ 1

]du = ln |u− 1| − ln |u+ 1|+ C = ln

√ex + 1− 1√ex + 1 + 1

+ C

68.1

x(x2 + x+ 1)=A

x+

Bx+ Cx2 + x+ 1

; A = 1, B = C = −1 so∫ −x− 1x2 + x+ 1

dx= −∫

x+ 1(x+ 1/2)2 + 3/4

dx = −∫u+ 1/2u2 + 3/4

du, u = x+ 1/2

= −12ln(u2 + 3/4)− 1√

3tan−1(2u/

√3) + C1

so∫

dx

x(x2 + x+ 1)= ln |x| − 1

2ln(x2 + x+ 1)− 1√

3tan−1 2x+ 1√

3+ C

69. u = sin−1 x, dv = dx, du =1√1− x2

dx, v = x;

∫ 1/2

0sin−1 x dx= x sin−1 x

]1/2

0−∫ 1/2

0

x√1− x2

dx =12sin−1 1

2+√1− x2

]1/2

0

=12

(π6

)+

√34− 1 = π

12+√32− 1

70.∫tan3 4x(1+ tan2 4x) sec2 4x dx =

∫(tan3 4x+ tan5 4x) sec2 4x dx =

116tan4 4x+

124tan6 4x+C

71.∫

x+ 3√(x+ 1)2 + 1

dx, let u = x+ 1,

∫u+ 2√u2 + 1

du=∫ [

u(u2 + 1)−1/2 +2√u2 + 1

]du =

√u2 + 1 + 2 sinh−1 u+ C

=√x2 + 2x+ 2 + 2 sinh−1(x+ 1) + C

Alternate solution: let x+ 1 = tan θ,∫(tan θ + 2) sec θ dθ=

∫sec θ tan θ dθ + 2

∫sec θ dθ = sec θ + 2 ln | sec θ + tan θ|+ C

=√x2 + 2x+ 2 + 2 ln(

√x2 + 2x+ 2 + x+ 1) + C.

January 27, 2005 11:45 L24-CH08 Sheet number 58 Page number 394 black

394 Chapter 8

72. Let x = tan θ to get∫

1x3 − x2 dx.

1x2(x− 1) =

A

x+B

x2 +C

x− 1 ; A = −1, B = −1, C = 1 so

−∫1xdx−

∫1x2 dx+

∫1

x− 1dx = − ln |x|+1x+ ln |x− 1|+ C

=1x+ ln

∣∣∣∣x− 1x∣∣∣∣+ C = cot θ + ln

∣∣∣∣ tan θ − 1tan θ

∣∣∣∣+ C = cot θ + ln |1− cot θ|+ C

73. lim�→+∞

− 12(x2 + 1)

]�a

= lim�→+∞

[− 12(.2 + 1)

+1

2(a2 + 1)

]=

12(a2 + 1)

74. lim�→+∞

1abtan−1 bx

a

]�0= lim�→+∞

1abtan−1 b.

a=

π

2ab